All language subtitles for College-Algebra-Full-Course_en

This video is about the exponent rules, rules\n

two to the fifth is just shorthand for two\n

five times. And similarly x to the n is just\n

when we write these expressions, the number\n

is called the base. And the number at the\n

the base by itself is called the exponent.

Sometimes the exponent is also called the\npower.

The product rule says that a 5x to the power\n

same thing as x to the n plus m power. In\n

For example, if I have two cubed times two\n

seventh. And that makes sense, because two\n

two by itself three times. And then I multiply\n

And in the end, I have to multiplied by itself\nseven times

I'm just adding up the number of times as\n

of times as multiplied total.

The quotient rule says that if I have x to\n

equal to x to the n minus m power. In other\n

then I can subtract their exponents.

For example, three to the six divided by three\n

two, or three to the fourth. And this makes\n

three by itself six times, and then I divide\n

So when I cancel out threes, I have four threes\nleft

notice that I have to subtract the number\n

threes at the top to get my number of threes\n

The power rule tells us if I have x to the\n

same thing as x to the n times M power. In\n

I get to multiply the exponents.

For example, five to the fourth cubed is equal\n

the 12th. And this makes sense, because five\n

to the fourth times five to the fourth, times\n

more, that's five times five times five times\n

So I have three groups of four, or five, which\n

The next rule involves what happens when I\n

power, it turns out that anything to the zeroeth\n

Usually, this is just taken as a definition.\n

If you have something like two cubed divided\n

equal one, anything divided by itself is just\n

we know that this is the same thing as two\n

divide two things with the same base, we get\n

is the same thing as two to the zero. So two\n

it work with the quotient rule.

And the same argument shows that anything\nto the zero power

What happens when we take something to a negative\npower

x to the n is equal to one over x to the n.\n

of a negative exponent. But here's why it\nmakes sense.

If I take something like five to seven times\n

roll the SAS to equal five to the seven plus\n

and we just said that that is equal to one.

Now I have the equation of five to the seventh\n

one, if I divide both sides by five, the seventh,\n

to equal one over five to the seventh. So\n

comes from. That has to be true in order to\n

Finally, let's look at a fractional exponents.\n

over N really mean? Well, it means the nth\n

means the cube root of 64, which happens to\n

square root of nine, which is usually written\n

Now, the square root of nine is just three.

fractional exponents also makes sense. For\n

cube that, then by the power role, that's\n

which is just five to the one or five. So\n

that when you cube it, you get five. And that's\n

the cube root of five is also a number that\n

The next rule tells us we can distribute an\n

In other words, if we have a product, x times\n

to x to the n times y to the N.

For example, five times seven

all raised to the third power is equal to\n

sense, because five times seven, all raised\n

times seven, times five times seven, times\n

of multiplication, this is the same thing\n

times seven times seven, or five cubed times\nseven cubed.

Similarly, we could distribute an exponent\n

over Y, all raised to the n power, that's\n

example, two sevenths raised to the fifth\n

over seven to the fifth. This makes sense,\n

as two sevens multiply by itself five times,\n

by itself five times divided by seven multiplied\n

fifth, over seven to the fifth, as wanted.

We've seen that we can distribute an exponent

over multiplication, and division.

But be careful, because we cannot distribute\nnext bowknot.

over addition, or subtraction, for example,\n

a to the n plus b to the n, a minus b to the\n

b to the n. And if you're not sure, just try\n

For example, two plus three squared is not\n

and two minus three squared is definitely\n

In this video, I gave eight exponent rules,\n

the quotient rule, the power rule

the zero exponent, the negative exponent,\n

and the two rules involving distributing exponents.

In another video, I'll use these exponent\n

In this video, I'll work out some examples\n

I'll start by reviewing the exponent rules.

The product rule says that when you multiply\n

the exponents. The quotient rule says that\n

base, you subtract the exponents. The power\n

power, you multiply the exponents. The power\n

power is one, as long as the base is not zero.

Since zero to the zero is undefined, it doesn't\nmake sense.

negative exponents to evaluate x to the minus\n

n. To evaluate a fractional exponent, like\n

we can distribute an exponent over a product,\n

times b to the n. And we can distribute an\n

is a to the n over b to the n.

In the rest of this video, we'll use these\n

For our first example, we want to simplify\n

x to the fourth, there's several possible\nways to proceed.

For example, we could use the negative exponent\n

all that gets divided by x to the fourth,\nstill

notice that we only take the reciprocal of\n

And that's because the exponent of negative\n

Now if we think of three as three over one,\n

numerator. And so we evaluate that by taking\n

of the denominators, which is three over x\n

I can think of x to the fourth as x to the\n

of a fraction, which I can evaluate by multiplying\n

times one divided by x squared times x to\n

using the product rule. Since x squared times\n

An alternate way of solving this problem

is to start by using the quotient rule

I can rewrite this as three times x to the\n

quotient rule, that's three times x to the\n

Now using the negative exponent roll, x to\n

this product of fractions simplifies to three\n

The second problem can be solved in similar\n

One way to simplify would be to use the negative\n

minus five as one over wide the fifth.

thinking of this as a fraction, divided by\n

and get four y cubed y to the fifth over one\n

four y to the eight. And so my final answer\n

Alternatively, I could decide to use the quotient\nrule first.

As in the previous problem, I can write this\n

quotient rule. And so that's for y to the\neighth as before.

I'd like to show you one more method to solve\n

To go on, that shortcut relies on the principle\n

corresponds to a positive exponent in the\n

two in the numerator here, after some manipulations\n

Furthermore, a negative exponent in the denominator

is equivalent to a positive exponent in the\nnumerator.

That's what happened when we had the y to\n

translated into a y to the positive five in\nthe numerator.

Sometimes people like to talk about this principle,\n

by switching the sine of the exponent that\n

Let's see how this principle gives us a shortcut\n

In the first problem, 3x to the minus two\n

exponent in the numerator and make it a positive\n

over x to the four plus two or x to the six.

In the second example, for y cubed over y\n

the minus five in the denominator into a y\n

answer of four y to the three plus five or\neight.

We'll use this principle again in the next\nproblems.

In this example, notice that I have I have\n

and also Z's in the numerator and the denominator.\n

get all my y's either in the numerator or\n

Since I have more y's in the denominator,\n

and make it a y to the negative three. I'm\n

in the numerator corresponds to a negative\n

Now since I have a positive exponent, z in\n

and I want to get rid of negative exponents,\n

as e to the minus two in the denominator,\n

Notice that my number seven doesn't move.\n

because it doesn't have an exponent, and the\n

applies to the Z not to the seven.

Now that I've got all my Z's in the numerator\n

to clean this up using the product rule.

And I have my simplified expression.

In this last example, we have a complicated\n

I'm going to start by simplifying the expression\n

I can bring all my y's downstairs and all\n

In other words, I can rewrite this as 25x\nto the fourth

I'll bring the Y to the minus five downstairs\n

bring the x to the minus six upstairs and\n

then I still have the Y cubed on the denominator,\n

Using the product rule, I can rewrite the\n

as 25x to the 10th over y to the eighth.

Recall that we're allowed to distribute an\n

When I distribute my three halves power

I get 25 to the three halves times x to the\n

Now the power rule tells me when I have a\n

So I can rewrite this as 25 to the three halves\n

y to the eight times three halves. In other\n

15th over y to the 12th. Finally, I need to\n

halves can be thought of as three times one\n

25 to the three halves as 25 to the three\n

Well, using the power rule in reverse, I can\n

or as 25 to the one half cubed. Since when\n

25 cubed to the one half might be hard to\n

but 25 to the one half is just the square\n

cubed, or five cubed, which is 125.

Therefore, my original expression is going\n

In this video, we use the exponent rules to\n

This video goes through a few tricks for simplifying\n

Recall that this notation means the nth root\n

root of eight, the number that when you cube\n

when we write the root sign without a little\n

In this case, the square root of 25 is five\n

Let's start by reviewing some rules for radical\n

of a product, we can rewrite that as the product\n

For example, the square root of nine times\n

nine times the square root of 16, you can\n

Similarly, it's possible to distribute a radical\n

by b is the same thing as the radical of A\n

the cube root of 64 over eight is the same\n

root of eight, and you can check that both\nof these evaluated to

you have to be a little bit careful though,\n

sign across addition. In general, the nth\n

of A plus the nth root of b. And similarly,\n

If you're ever in doubt, you can always check\n

root of one plus one is not the same thing\n

the right side evaluates to one plus one or\n

The second expression to show that that fails,\n

root of one minus one it'll actually hold\n

using say the square root of two minus one,\n

You might notice that these Rules for Radicals,\n

hold, remind you of rules for exponents. And\n

be written in terms of exponents. For example,\n

this, the nth root of A times B is the same\n

exponent rules, I can distribute an exponent\n

rule can be restated completely in terms of\n

can be restated in terms of exponents as a\n

one over n divided by b to the one over n.\n

and the exponents to rewrite a to the m over\n

a to the m with the N through taken. That's\nalso the same thing

As the nth root of A, all taken to the nth\n

exponent rules. So a to the m over N is the\n

over nth power. That's because when we take\n

and M times one over n is equal to M over\nn.

But a one over nth power is the same thing\n

is the same thing as this expression. And\n

equivalence can we prove similarly, by writing\n

M. Again, this is works because when I take\n

one over n times M is the same thing as M\n

the same, because the one over nth power is\n

One mnemonic for remembering these relationships\n

and root is like root, so that tells us we\n

the power, and the n becomes the root in either\n

Now let's use these rules in some examples.\n

halves power, well, first I'll use my exponent\n

one over 25 to the three halves power.

Next, I'll use the power of a root mnemonic\n

rooted, or, as 25 square rooted to the third\npower.

I wrote the two's there for the square root\n

will omit this and just write the square root\n

Now, I could use either of these two equivalent\n

this one because it's easier to compute without\n

five, five cubed is 125. So my answer is one\nover 125.

If I tried to compute the cube of 25, first,\n

easier to compute the route before the power\n

Now let's do an example simplifying a more\n

cynet. I want to take the square root of all\n

negative exponents, I'm first going to rewrite\n

the sixth over z to the 11th. So I'll change\n

by moving this, this factor to the denominator.

Now, when you're asked to simplify radical\n

much as possible, out of the radical side.

To pull things out of the square root side,\n

rewrite everything in terms of squares as\n

a square, those two operations undo each other.\n

60. So 60 is going to be two squared times\n

Now I'll break things up into squares as much\n

three times five, I've already got an x squared,\n

y squared times y squared. And I'll write\n

I guess five times 12345 times when extra\n

add all those exponents together.

Now I know that I can distribute my radical\n

I'll write this with a zillion different radicals\nhere.

And every time I see the square root of something\n

and the squares out and get what's what's\n

I get two times the square root of three times\n

z times itself, I guess five times times the\n

up with exponents. I'll write that as the\n

times x times y cubed over z to the fifth\nthe square root of z.

I'm gonna leave this example as is. But sometimes\n

without radical signs in the denominator.\n

I won't do it here, but I'll show you how\n

This example asks us to rationalize the denominator,\n

without radical signs in the denominator.

To get rid of the radical sine and the denominator,\n

root of x. But I can't just multiply the denominator\n

the numerator by the same thing. So then just\n

and I don't change the value of my expression.

Now, if I just multiply together numerators\n

squared of x 10 squared of x is the square\n

Now I can cancel my access from the numerator\n

times the square root of x. I rationalize\n

In this video, we went over the Rules for\n

by working with fractional exponents

pulling things out of the radical sign

and rationalizing the denominator.

This video goes over some common methods of\n

means to write it as a product. So we could\n

times five, we could factor it more completely\n

As another example, we could factor the expression\n

x plus two times x plus three.

In this video, I'll go over how I get from\n

But for right now, I just want to review how\n

factoring is correct. And that's just by multiplying\n

two times x plus three, then I multiply x\n

gives me 3x. Two times x gives me 2x. And\n

to x squared plus 5x plus six, which checks\n

of factoring as the opposite of distributing\n

by distributing or multiplying out

a bit of terminology, when I think of an expression\n

I sum up are called the terms. But if I think\n

then the things that I multiply together are\ncalled factors.

Now let's get started on techniques of factoring.\n

like to start by pulling out the greatest\n

means the largest thing that divides each\nof the terms.

In this first example, the largest thing that\n

So the GCF is five. So I pull the five out,\n

number. And so I get three plus 5x.

Pause the video for a moment and see if you\n

The biggest thing that divides both x squared\n

One way to find this is to look for the power\n

So that's x squared. And the power of y that\n

Now if I factor out the x squared y from each\n

by x squared y, if I divide the first term\n

the second term by x squared y, I'm going\n

out on the side just to make it more clear,\n

Three x's on the top and two x's and a y on\n

a Y. So I'll write the x, y here, and I factored\n

by multiplying out. So if I multiply out my\n

the first term and the second term, I get\n

together and two y's multiplied together.\n

The next technique of factoring, I'd like\n

example, notice that we have four terms, factoring\n

If you have four terms in your expression,\nyou need to factor

in order to factor by grouping, I'm first\n

of the first two terms, and then separately,\n

last two terms. The greatest common factor\n

I factor out the x squared, and I get x plus\n

of forex and 12 is just four. So I factor\n

Notice that the factor of x plus three now\n

the greatest common factor of x plus three,\n

of the right. And now I have an x squared\n

this second piece. And that completes my factoring\n

factor further by factoring the expression\n

see later, this expression, which is a sum\n

does not factor any further over the integers.

Next, we'll do some factoring of quadratics.\n

just a term with x in it, and a constant term\nwith no x's in it.

I'd like to factor this expression as a product\n

The key idea is that if I can find those two\n

this expression, those two numbers would have\n

eight. And these two numbers would end up\n

because when I multiply out, this number will\n

be also another coefficient of x, they'll\n

So if I look at all the pairs of numbers that\n

could be one and eight, two, and four, four\n

as I had before. And that's sort of the same\n

negatives, I could have negative one, negative\n

four, those alternate multiply together to\n

if there's a pair of these numbers that add\n

that these ones will work. So now I can write\n

two times x minus four. And it's always a\n

going to get x squared minus 4x minus 2x,\n

I want. Now this second examples a bit more\n

my coefficient of x squared is not just one,\nit's the number 10.

Now, there are lots of different methods for\n

going to show you one method, my favorite\n

but to start out, I'm going to multiply my\n

so I'm multiplying 10 by negative six, that\n

coefficient of x the number 11. And write\n

two numbers that multiply to give me negative\n

that this is exactly what we were doing in\n

have to multiply the coefficient of x squared\n

was just one. So to find the two numbers that\n

might just be able to come up with them in\n

can figure it out. Pretty simple.

thematically by writing out all the factors,\n

60. So I can start with negative one and 60,\n

And keep going like this until I have found\n

me the number 11. And, and now that I look\n

gives me 11. So I don't have to continue with\n

factors, I write out my expression 10x squared,\n

4x plus 15x. Now I copy down the negative\n

11x. That's how I chose those numbers. And\n

as, as this expression, I haven't changed\n

that I can apply factoring by grouping on\n

I factor out my greatest common factor of\n

it's 2x. So I factor out the 2x, I get 5x\n

common factor of 15x and negative six, that\n

again, this is working beautifully. So I have\n

the 5x minus two on the right, and I put what's\n

plus three. And I have factored my expression.

There are a couple special kinds of expressions\n

just memorize the formula for. So the first\n

something of the form a squared minus b squared,\n

a minus b. And let's just check that that\n

multiply that out, I get a squared minus a\n

two terms cancel out. So it gives me back\n

it. So for this first example, I if I think\n

squared, then I can see that's a difference\n

as x plus four times x minus four. And the\n

that's the same thing as three p squared minus\n

Notice that if I have a sum of squares

x squared plus four, which is x squared plus\n

The difference of squares formula doesn't\n

for a sum of squares. There is, however, a\n

a sum of cubes. The difference of cubes formula,\n

squared plus a b plus b squared. The formula\n

you just switch the negative and positive\n

b times a squared minus a b plus b squared.\n

multiplying out. Let's look at one example\n

actually a sum of two cubes because it's y\n

using the sum of cubes formula by plugging\n

y plus three times y squared minus y times\n

that up a little bit to read y plus three\n

So in this video, we went over several methods\n

common factor. We did factoring by grouping.\n

difference of squares. And we did a difference\n

and more complicated problems, you may need\n

to get through a single problem. For example,\n

Add to a factoring of quadratics, or something\nsimilar.

This video gives some additional examples\n

which of these first five expressions factor\n

The first expression can be factored by pulling\n

So that becomes x times x plus one.

The second example can be factors as a difference\n

something squared, minus something else squared.\n

like a squared minus b squared, that's a plus\n

X plus five times x minus five.

The third one is a sum of two squares, there's\n

real numbers. So this is the one that does\nnot factor.

just for completeness, let's look at the next\n

When we factor by grouping, we pull up the\n

terms, that would be an x squared, that becomes\n

as much as we can add the next two terms,\n

that the x plus two factor now occurs in both\n

x plus two out and get x plus two times x\nsquared plus three

we can't factor any further because x squared\n

Finally, we have a quadratic, this also factors.\n

by grouping trick. So first, what I do is\n

the constant term, five times eight is 40.\n

take the coefficient of the x term, that's\n

part of the x. Now I'm looking for two numbers\n

Sometimes I can just guess numbers like this,\n

40. So factors of 40, I could do one times\n

add to a negative number. So if I use two\n

to add to a negative number. It's better for\n

a negative times a negative still multiplies\n

a negative number. So but negative one and\n

add to negative 14, they add to negative 41.\n

biggest number that divides 40, besides one\n

20. Those add to negative 22. That doesn't\n

so I'll try negative four and negative 10.\n

negative 10 is negative 14, negative four\n

it. Alright, so the next step is to use factoring\n

this negative 14x as negative 4x minus 10x.\n

Notice that this works, because I picked negative\n

14, so so negative 4x minus 10x, will add\n

just just expand it out a little bit. Now\n

so I can group the first two terms and factor\n

x times 5x minus four. And now I'll factor\n

including the the negative. So that becomes,\n

and that becomes 5x minus four since negative\n

I've got the same 5x plus four in both my\n

I can factor out the 5x minus four from both\n

I factored this quadratic. If I want to, of\n

out by multiplying out. So a check here would\n

squared 5x minus 10 to minus two is minus\n

four times minus two is plus eight. So let's\n

should be. So that was the method of factoring\na quadratic.

And all of these factors except for the sum\nof squares.

So we saw that factoring by grouping is handy\n

also handy for factoring the quadratic indirectly,\n

into two terms. So how can you tell when a\n

by grouping, there's, there's an easy way\n

that's that it has four terms.

So if you see four terms, or in the case of\n

then that's a good candidate for factoring\n

two terms group the second two terms, factoring\n

with four terms. But, but that's like the\n

what are the same main techniques of factoring.\n

there was pull out common factors. There's\n

There's factoring quadratics.

And one more that I didn't mention is factoring\n

that uses the formulas, aq minus b cubed is\n

squared. And a cubed plus b cubed is a plus\n

One important tip when factoring

I always recommend doing this first, pull\n

That'll simplify things and making the rest\n

you might need to do several these factoring\n

you might have to first pull out a common\n

And then you might notice that one of your\n

and you have to apply a difference of squares\n

bit, keep factoring as far as you can go.

Here are some extra examples of factoring\n

the video and give these a try.

For the first one, let's multiply two times\n

then we'll bring the three down in the bottom\n

that multiply to negative 28 and add to three.\n

28, we'll need one of them to be negative\n

negative 128, or one, negative 28, those don't\n

14, those don't work. Hey, I just noticed\n

the negative number, so they add to a positive\n

negative four times seven, four times negative\n

will work. So I'll write those here at negative\nfour, seven

copy down the two z squared. And I'll split\n

z and then minus 14. Now factoring by grouping,\n

pull out a seven and that becomes z minus\n

seven times z minus two as my factored expression.

My second expression, I could work at the\n

grouping that kind of thing. But it's actually\n

I can pull out a common factor from all of\n

simpler to deal with. So notice that a five\n

I'm going to go ahead and pull out the negative\n

in front of my squared term. So I'm going\n

Again, it would work if I forgot to do this,\n

negative five v squared, this becomes minus,\n

negative five is negative 45. And this becomes\nminus

10 cents native 10 times negative five is\n

by grouping, or I can use kind of a shortcut\n

I can just put these here, and then I know\n

have to multiply to the negative 10. And they're\n

be plus 10, and a minus one will do the trick.

Those are all my factoring examples for today.\n

a chance to spend some time working in ALEKS.\nBye.

This video is about working with rational\n

usually with variables in it, something like\n

rational expression. In this video, we'll\n

and dividing rational expressions and simplifying\n

We'll start with simplifying to lowest terms.\n

numbers in it, something like 21 over 45,\n

and then canceling common factors.

So in this example, the three is cancel, and\n

If we want to reduce a rational expression\n

we proceed the same way. First, we'll factor\n

and then factor the denominator. In this case\n

could also write that as x plus two squared.\n

left with three over x plus two. Definitely\n

Next, let's practice multiplying and dividing.\n

just numbers in them, we simply multiply the\n

So in this case, we would get four times two\n

If we want to divide two fractions, like in\n

as multiplying by the reciprocal of the fraction\n

times three halves, and that gives us 12 tenths.\n

we use the same rules when we compute the\n

with the variables. And then here, we're trying\n

we can multiply by the reciprocal. I call\n

And now we just multiply the numerators.

And multiply the denominators.

It might be tempting at this point to multiply\n

denominator. But actually, it's better to\n

even more completely. That way, we'll be able\n

the common factors. So let's factor even more\n

plus one, and x squared minus 16. And that's\n

four times x minus four, the denominator is\n

it over. And now we can cancel common factors\n

x minus four. This is our final answer.

Adding and subtracting fractions is a little\n

find a common denominator. A common denominator\n

into, it's usually best of the long run to\n

the smallest expression that both denominators\ndivided into.

In this example, if we just want a common\n

is 90 because both six and 15 divided evenly\n

the best way to do that is to factor the two\n

times five, and then put together only the\nfactors we need for

Both six and 50 into divider numbers. So if\n

is 30, we know that two times three will divide\n

it. And we won't be able to get a denominator\n

three, and five, in order to ensure both these\n

denominator, we can rewrite each of our fractions\n

I need to get a 30 in the denominator, so\n

and multiply by the factors that are missing\n

my least common denominator of 30. For the\n

30. So I'm going to multiply by two over two

I can rewrite this as 3530 s minus 8/30. And\n

just subtract my two numerators. And I get\n27/30.

If I factor, I can reduce this

to three squared over two times five, which\n

sum of two rational expressions with variables\n

we have to find the least common denominator,\n

So 2x plus two factors as two times x plus\n

of two squares. So that's x plus one times\n

I'm going to take all the factors, I need\n

into, so I need the factor two, I need the\n

minus one, I don't have to repeat the factor\n

And so I will get my least common denominator\n

not going to bother multiplying this out,\n

form to help me simplify later. Now I can\n

by multiplying by whatever's missing from\n

denominator. So what I mean is, I can rewrite\n

plus two is two times x plus one, I'll write\n

compared to the least common denominator,\n

I multiply the numerator and the denominator\n

But I can't get away with just multiplying\n

I have to multiply by it on the numerator\n

by one and a fancy form and not changing the\n

the second rational expression, I'll I'll\n

make it easier to see what's missing from\n

compared to my least common denominator is\n

and the denominator by two. Now I can rewrite\n

becomes three times x minus one over two times\n

five times two over two times x plus 1x minus\n

So I can just add together my numerators.\n

two times x plus 1x minus one. I'd like to\n

is to leave the denominator in factored form.\n

so that I can add things together. So I get\n

1x minus one, or 3x plus seven over two times\n

factor. And there's therefore no factors that\n

As much as it can be. This is my final answer.

In this video, we saw how to simplify rational\n

We also saw how to multiply rational expressions\n

the denominator, how to divide rational expressions\n

and subtract rational expressions by writing\n

This video is about solving quadratic equations.\n

the square of the variable, say x squared,\n

The standard form for a quadratic equation\n

represent real numbers. And a is not zero\n

Let me give you an example. 3x squared plus\n

in standard form, here a is three, B is seven,\n

equals minus 7x plus two is also a quadratic\n

The key steps to solving quadratic equations\n

form, and then either factor it

or use the quadratic formula, which I'll show\n

Let's start with the example y squared equals\n

and we need to rewrite this quadratic equation\n

18 from both sides and adding seven y to both\n

minus 18 plus seven y equals zero. And I can\n

seven y minus 18 equals zero. Now I've got\n

to try to factor it. So I need to look for\n

two numbers that work are nine and negative\n

left as y plus nine times y minus two equals\n

that multiply together to give you zero, either\n

second quantity has to be zero, or I suppose\n

this is really handy, because that means that\n

or y minus two equals zero. So I can as my\n

y plus nine equals zero, or y minus two equals\n

It's not a bad idea to check that those answers\n

equation, negative nine squared, does that\n

and you can work out that it does. And similarly,\n

In the next example, let's find solutions\n

is a quadratic equation, because it's got\n

in standard form by subtracting 121 from both\nsides.

Notice that A is equal to one b is equal to\n

to negative 121 in the standard form, a W\n

I'm going to try to factor this expression.\n

of two squares, and it factors as w plus 11,\n

I get w plus 11 equals zero or w minus 11\n

11. In this example, I could have solved the\n

said that if W squared is 121, and then w\n

of 121. In other words, W is plus or minus\n11.

If you saw the equation this way, it's important\n

11 squared equals 121, just like 11 squared\ndoes.

Now let's find the solutions for the equation.\n

might be tempted to say that, oh, if two numbers\n

better equal one and the other equals seven\n

But that's faulty reasoning in this case,\n

whole numbers. They could be crazy.

fractions or even irrational numbers. So instead,\n

To do that, I'm first going to multiply out.\n

That equals seven, and I'll subtract the seven\n

seven is zero. Now I'm looking to factor it.\n

seven and add to two, since the only way to\n

seven or seven times negative one, it's easy\n

will do will work. So there's no way to factor\n

let's use the quadratic equation. So we have\n

So A is one, B is two, and C is minus seven.\n

quadratic equation, which goes x equals negative\n

minus four, I see all over two

different people have different ways of remembering\n

it x equals negative b plus or minus the square\n

to a, but you can use any pneumonic you like.\n

negative two plus or minus the square root\n

negative seven support and remember the negative\n

Now two squared is four, and four times one\n

whole quantity under the square root sign\n

I can rewrite this as x equals negative two\n

Since 32, is 16 times two and 16 is a perfect\n

plus or minus the square root of 16 times\n

negative two plus or minus four times the\n

Next, I'm going to split out my fraction

as negative two over two plus or minus four\n

those fractions. This becomes negative one\n

answers are negative one plus two square root\n

of two. And if I need a decimal answer for\n

As our final example, let's find all real\n

equals 1/3 y minus two. I'll start as usual\n

me one half y squared minus 1/3 y plus two\n

to factor or use the quadratic formula right\n

of annoying. So I'd like to get rid of them.\n

that means I'm going to multiply the whole\n

In this case, the least common denominator\n

the whole equation by six have to make sure\n

in this case, six times zero is just zero.\n

y squared minus two y plus 12 equals zero.

Now, I could try to factor this, but I think\n

So I get x equals negative B, that's negative\n

root of b squared, minus four times a times\nc, all over to a.

Working out the stuff in a square root sign,\n

So this simplifies to x equals to plus or\n

negative 140. All of our sex. Well, if you're\n

the square root sign, you should be we can't\n

and get an N get a real number is our answer.\n

number. And therefore, our conclusion is we\n

In this video, we solve some quadratic equations\n

Either factoring or using the quadratic formula.

In some examples, factoring doesn't work,\n

But in fact, using the quadratic formula will\n

solve it by factoring. So you can't really\n

just sometimes it'll be faster to factor instead.

This video is about solving rational equations.\n

that has rational expressions in that, in\n

There are several different approaches for\n

start by finding the least common denominator.\n

three and x, we can think of one as just having\n

Since the denominators don't have any factors\n

just by multiplying them together.

My next step is going to be clearing the denominator.

By this, I mean that I multiply both sides\n

x plus three times x, I multiply on the left\n

same thing on the right side of the equation.

Since I'm doing the same thing to both sides\n

of the equation. Multiplying the least common\n

is equivalent to multiplying it by all three\n

I'll rewrite the left side the same as before,\n

right side to get x plus three times x times\n

x. So I've actually multiplied the least common\n

Now I can have a blast canceling things. The\n

The here are nothing cancels out because there's\n

numerator cancels with the x in the denominator.

So I can rewrite my expression as x squared\n

x plus three. Now I'm going to simplify.

So I'll leave the x squared alone on this\n

plus x plus three, hey, look, the x squared\n

equals 4x plus three, so 4x is negative three,\n

I'm going to plug in my answer to check. This\n

it's especially important for a rational equation\n

you'll get what's called extraneous solution\n

original equation because they make the denominator\n

we're going to get any extraneous equations\n

to make any of these denominators zero, so\n

the denominator here, negative three fourths\n

nine fourths. And this is one I'll flip and\n

I can simplify my complex fraction, it ends\n

four thirds is negative 1/3. So that all seems\nto check out.

And so my final answer is x equals negative\nthree fourths.

This next example looks a little trickier.\n

First off, find the least common denominator.\n

c plus one, and C squared minus four c minus\n

five times c plus one. Now, my least common\n

to that each of these denominators divided\n

I need the factor c plus one. And now I've\n

denominator. So here is my least common denominator.\n

So I do this by multiplying both sides of\n

In fact, I can just multiply each of the three\n

I went ahead and wrote my third denominator\n

what cancels. Now canceling time dies, this\n

out the denominator is the whole point of\n

you're multiplying by something that's big\n

you don't have to deal with denominators anymore.

Now I'm going to simplify by multiplying out.

So I get, let's see, c plus one times four\n

I get minus just c minus five, and then over\n

I can rewrite the minus quantity c minus five\n

And now I can subtract the three c squared\n

here, and the four c minus c that becomes\na three C.

And finally, I can subtract the three from\n

two equals zero. got myself a quadratic equation\n

this factors to C plus one times c plus two\n

or C plus two is zero. So C equals negative\n

Now let's see, we need to still check our\nanswers.

Without even going to the trouble of calculating\n

one is not going to work, because if I plug\n

zero, which doesn't make sense. So C equals\n

actually satisfy my original equation. And\n

I can go if I go ahead, and that doesn't make\n

made any mistakes, it should satisfy my original\n

So my final answer is C equals negative two.\n

equations using the method of finding the\n

we cleared the denominator by multiplying\n

denominator or equivalently. multiplying each\n

There's another equivalent method that some\n

we find the least common denominator, but\n

over that least common denominator. So in\n

denominator of x plus three times x. But our\n

rational expressions over that common denominator\n

appropriate things. So one, in order to get\n

to multiply the top and the bottom by x plus\n

the top and the bottom just by x plus three\n

x. Now, if I simplify a little bit

let's say this is x squared over that common\ndenominator, and

here I have just x plus three times x over\n

over that common denominator. Now add together\n

So this is x plus three times x plus x plus\n

that are equal that have the same denominator,\n

also. So the next step is to set the numerators\nequal.

So I get x squared is x plus three times x\n

the previous way, we solve this equation,\n

here on we just continue as before.

When choosing between these two methods, I\npersonally tend

prefer the clear the denominators method,\n

don't have to get rid of those denominators\n

times. But some people find this one a little\n

understand either of these methods is fine.

One last caution, don't forget at the end,\n

These will be solutions that make the denominators\n

This video is about solving radical equations,\n

square root signs in them, or cube roots or\n

When I see an equation with a square root\n

root. But it'll be easiest to get rid of the\n

root. In other words, I want to get the term\n

of the equation by itself, and everything\n

I start with my original equation, x plus\n

x from both sides, then that does isolate\n

everything else on the right. Once I've isolated\n

rid of the square root. And I'll do that by\nsquaring both sides

of my equation. So I'll take the square root\n

Now the square root of x squared is just x,\n

to work out 12 minus x squared, write it out\n

minus x is minus 12x. I get another minus\n12x from here.

And finally minus x times minus x is positive\n

that's minus 24x. And now I can subtract x\n

25x plus x squared. That's a quadratic equation,\n

So now I've got a familiar quadratic equation\n

I'll just proceed to solve it like I usually\n

So I'm going to look for two numbers that\n

I'm going to need negative numbers to get\n

negative numbers. So they still multiply to\n

factors of 144, I could have negative one\n

72, negative four, negative 36, and so on.\n

it's not hard to find the two that add to\n

16. So now I can factor in my quadratic equation\n

that means that x minus nine is zero or x\n

16. I'm almost done. But there's one last\n

Solutions so that we can eliminate any extraneous\n

that we get that does not actually satisfy\n

can happen when you're solving equations with\n

nine. If we plug in to our original equation,\n

want that to equal 12. Well, the square root\n

indeed equal 12. So that solution checks out.\n

get 16 plus a squared of 16. And that's supposed\n

is supposed to equal 12. But that most definitely\n

to be extraneous solution, and our only solution\n

This next equation might not look like an\n

we can think of a fractional exponent as being\n

the same thing we did on the previous problem\n

that involves the fractional exponent. So\n

times P to the four fifths equals 1/8. And\n

I can multiply both sides by one half, that\n

And I've effectively isolated the part of\n

as much as possible. Now, in the previous\n

radical. In this example, we're going to get\n

to actually do this in two stages. First,\n

That's because when I take an exponent to\n

so that becomes just p to the fourth equals\n

get rid of the fourth power by raising both\n

root, there's something that you need to be\n

an even root, or the one over an even number\n

It's kind of like when you write x squared\n

of both sides, x could equal plus or minus\n

plus or minus two, since minus two squared\n

why when you take an even root, or a one over\n

to include the plus or minus sign, when it's\n

don't need to do that. If you had something\n

equals the cube root of negative eight, which\n

you don't need to do the plus or minus because\n

So that aside, explains why we need this plus\n

1/4. When I raise a power to a power, I multiply\n

which is equal to plus or minus 1/16 to the\n

Now I just need to simplify this expression,\n

power, because 16 to the fifth power is like\n

as P equals plus or minus 1/16. I'll write\n

And as I continue to solve using my exponent\nrules

I'm going to prefer to write this as 1/16\n

it's going to be easier to take the fourth\n

the same thing as the fourth root of one over\n

fifth power. fourth root of one is just one\n

to the fifth power, that's just going to be\n

The last step is to check answers.

So I have the two answers p equals 130 seconds,\n

1/32 to the fourth fifth power.

That gives me two times one to the fourth\n

two times one over 32/5 routed to the fourth\n

Raisa to the fourth power, I get 16. So this\n

wanted in the original equation up here. Similarly,\n

actually does satisfy the equation, I'll leave\n

So our two solutions are p equals one over\n

to point out an alternate approach to getting\n

gotten rid of it all in one fell swoop by\n

five fourths is the reciprocal of four fifths.\n

when I raise the power to the power, I multiply\n

fifths times five fourths is plus or minus\n

P to the One Power, which is just P is plus\n

an alternate and possibly faster way to get\n

comes from the fact that when we take the\n

a fourth root, and so we need to consider\n

This video is about solving radical equations,\n

square root signs in them, or cube roots or\n

In this video, we solved radical equations\n

and then removing the radical sine or the\nfractional exponent

by either squaring both sides or taking the\n

This video is about solving equations with\n

Recall that the absolute value of a positive\n

value of a negative number is its opposite.\n

of a number as representing its distance from\n

and the number of negative four are both at\n

value of both of them is four.

Similarly, if I write the equation, the absolute\n

to be three units away from zero on the number\nline.

And so X would have to be either negative\n

three times the absolute value of x plus two\nequals four.

I'd like to isolate the absolute value part\n

subtracting two from both sides

and dividing both sides by three.

Now I'll think in terms of distance on a number\n

means that x is two thirds away from zero.\n

And the answer to my equation is x is negative\n

I can check my answers by plugging in

three times the added value of negative two\n

four. Well, the absolute value of negative\n

three times two thirds plus two, which works\nout to four.

Similarly, if I plug in positive two thirds,\n

The second example is a little different,\n

a more complicated expression, not just around\nthe X.

I would start by isolating the absolute value\npart.

But it's already isolated. So I'll just go\n

on the number line. So on my number line,\n

to be at a distance of four from zero.

So that means that 3x plus two is here at\nfour or 3x plus two

though is it negative four, all right, those\nas equations

3x plus two equals four, or 3x plus two is\n

So this becomes 3x equals two, or x equals\n

minus six, or x equals minus two. Finally,\n

I'll leave it to you to verify that they both\nwork.

A common mistake on absolute value equations\n

like we did here, and then just solve for\n

Another mistake sometimes people make is,\n

assume that the negative of that works also.

But that doesn't always work. In the first\n

of each other. But in our second examples,\n

of each other one was two thirds and the other\nwas negative two.

In this third example, let's again, isolate\n

So starting with our original equation

we can subtract 16 from both sides

and divide both sides by five or equivalently,\nmultiply by 1/5.

Now let's think about distance on the number\nline

we have an absolute value needs to equal negative\n

absolute value sign needs to be at distance\n

be at distance negative three from zero. Another\n

the absolute value of something and end up\n

positive, or zero. So this equation doesn't\n

In this video, we solved absolute value equations.\n

will have two solutions. But in some cases,\n

This video is about interval notation, and\n

Before dealing with interval notation, it\n

Our first example of an inequality is written\n

First, we used to write down our variable\n

values for this problem that is one and three.\n

numbers, meaning we will have one inequality\n

it says not including one and three, which\n

beneath each inequality. Here we will put\n

put three the highest key value. Next, we're\n

Here we write our key values one, and three.\n

we have an empty circle around each number.\n

The last step of this problem is writing this\n

writing things in interval notation is kind\n

put one and then here we will put three.

Next we need to put brackets around these\nnumbers.

For this problem, it is not including one\n

However, if it were including one and three,\n

It is important to note for interval notation,\n

left and the biggest value always goes on\nthe right.

You also include a comma between your two\nkey values.

Now let's work on problem B. Once again we\n

negative four and two but this time it is\n

us to have the or equal to assign below the\n

here and their highest here. The number line\n

have our key values negative four and two.\n

right here, we instead use a closed circle,\n

four and two, you complete this with a line\n

notation. In the last problem, I said that\n

we would use these hard brackets. Now we are\n

on each side because isn't including foreign\n

left, and our highest value on the right,\n

You now know how to correctly write these\n

Now we are going to practice transforming\n

notation, and vice versa. This is slightly\n

of having two soft brackets or two hard brackets,\n

left side will have a hard bracket because\n

right side, it is a soft bracket because it\n

put a comma in the middle, a lower key value\n

we're taking an equation already written an\n

For the second problem, we can see are key\n

Now, this sounds a little bit weird, but let's\n

x, which is less than because of the soft\n

or equal to because it has a soft bracket\n

greater than negative infinity, we can take\n

It is important to note that a soft bracket\n

infinity is not a real number, so we cannot\n

The next problem is a little tricky, because\n

But let's just start off the equation as a\n

of an order equal to sign. And negative 15\n

On the other side, we put the highest possible\n

a soft bracket. We can see the relation of\n

into this problem when x is greater than a\n

Once again, we have our sauce bracket for\n

Part D brings up an important point about\n

our other problems, we have had the inequalities\npointing left

instead of rights. When seeing Part D, you\n

would go here, and zero is on the right who\n

When writing inequalities, you must always\n

this problem is zero. To fix this, we can\n

this, it is still identical just written in\n

notation, then we can see that zero has a\n

zero, but r comma four or other key value\n

For interval notation, you must always have\n

This was our video on interval notation, an\n

This video is about solving inequalities that\n

Let's look at the inequality absolute value\n

Thinking of absolute value as distance. This\n

is less than five units. So x has to live\n

We can express this as an inequality without\n

is less than x, which is less than five. Or\n

negative five, five, soft bracket.

Both of these formulations are equivalent\n

In the second example, we're looking for the\n

x is greater than or equal to five.

on the number line, this means that the distance\n

A distance bigger than five units means that\n

over here, where it's farther than five units\n

distance equal to five units. So I'll fill\n

the number line that satisfy my inequality.

Now I can rewrite the inequality without the\n

less than or equal to negative five, or x\n

also write this in interval notation

soft bracket, negative infinity, negative\n

hard bracket five infinity soft bracket, I\n

I'm trying to describe all these points on\n

Let's take this analysis a step further with\n

want the absolute value of three minus two\n

And an absolute value less than four means\n

But it's not the variable t that lives in\n

zero, it's the whole expression, three minus\ntwo t.

So three minus two t, live somewhere in here.\n

absolute value signs by saying negative four\n

four. Now I have a compound inequality that\n

three from all three sides to get negative\n

than one. And now I'll divide all three sides\n

number, this reverses the directions of the\ninequalities.

Simplifying, I get seven halves is greater\n

So my final answer on the number line looks\n

But not including the endpoints, and an interval\n

a half, seven, half soft bracket

please pause the video and try the next problem\non your own.

thinking in terms of distance, this inequality\n

three minus two t and zero is always bigger\n

If three minus two t has a distance bigger\n

region that's near zero, it has to be on the\n

That is three minus two t is either less than\n

than four. I solve these two inequalities\n

from both sides, I get negative two t is less\n

I get T is bigger than seven halves. And then\n

greater than one. So t is less than negative\n

last step due to dividing by a negative number.\n

again, the first piece says that t is greater\n

And the second piece says that t is less than\n

Because these two statements are joined with\n

in this one, or in this one. That is I want\n

interval notation, this reads negative infinity\n

This last example looks more complicated.\n

value part, it looks pretty much like the\n

seven from both sides. And then I'll divide\n

And I'm looking for this expression for x\n

than or equal to negative three from zero,\n

to negative three, well, distance is always\n

always greater than equal to zero. So this,\n

And so the answer to my inequality is all\n

In other words, all real numbers.

Once solving absolute value inequalities,\n

value of something that's less than a number\n

On the other hand, an absolute value is something\n

whatever's inside the absolute value sign\n

from zero is bigger than that certain number.

drawing these pictures on the number line\n

and equality as an inequality that doesn't\n

it would be negative three is less than x\n

case, it would be either x plus two is less\n

This video is about solving linear inequalities.\n

involve a variable here x, but don't involve\n

The good news is, we can solve linear inequalities,\n

adding and subtracting terms to both sides\n

both sides. The only thing that's different\n

number, then you need to reverse the direction\n

the inequality, negative x is less than negative\n

by negative one to get rid of the negative\n

the inequality. With this caution in mind,\n

Since our variable x is trapped in parentheses,\n

That gives me negative 5x minus 10 plus three\n

Negative 10 plus three is negative seven,\n

seven is greater than eight. Now I'll add\n

greater than 15. Now I'd like to divide both\n

is a negative number that reverses the inequality.\n

five. In other words, x is less than negative\nthree.

If I wanted to graph this on a number line,\n

an open circle around it, and shade in to\nthe left

I use an open circle, because x is strictly\n

three. If I wanted to write this in interval\n

infinity, negative three soft bracket. Again,\n

is not included. This next example is an example\n

Either this statement is true, or this statement\n

satisfy either one. I'll solve this by working\n

them at the end. For the inequality on the\n

then subtract x from both sides

and then divide both sides by two.

I didn't have to reverse the inequality sign\n

on the right side, I'll copy the equation\n

both sides by 696 is the same as three halves.\n

row four, make this statement true. Let me\n

x is less than or equal to negative two, means\n

to the left. X is greater than three halves\n

My final answer includes both of these pieces,\n

allowed to be an either one or the other.\n

The first piece on the number line can be\n

negative two hard bracket. And the second\n

halves, infinity soft bracket to indicate\n

I use the union side, which is a U.

That means that my answer includes all x values\n

This next example is also a compound inequality.\n

I'm looking for all y values that satisfy\n

I can solve each piece separately

on the left, to isolate the Y, I need to multiply\n

that gives me Why is less than negative 12\n

flip to a less than because I was multiplying\n

By clean up the right side, I get why is less\nthan 18.

On the right side, I'll start by subtracting\n

And now I'll divide by negative four, again,\n

So that's why is less than three over negative\n

three fourths. Again, I'm looking for the\n

Let me graph this on the number line

the Y is less than 18. I can graph that by\n

I don't want to include it. So I use an empty\n

the statement y is less than negative three\n

And again, I don't include it, but I do include\n

values for which both of these statements\n

both colored blue and colored red. And so\n

draw it above so you can see it easily. So\n

three fourths and lower, not including negative\n

the number line that have both red and blue\n

answer will be soft bracket negative infinity\n

As my final example, I have an inequality\n

three is less than or equal to 6x minus two\n

a compound inequality with two parts, negative\n

And at the same time 6x minus two is less\n

before. But instead, it's a little more efficient\n

same thing to all three sides. So as a first\n

gives me negative one is less than or equal\n

divide all three sides by six to isolate the\n

than or equal to x is less than two.

If we solved it, instead, in two pieces above,\n

get negative one six is less than or equal\n

than two on this piece. And because of the\n

negative one, six is less than or equal to\n

Either way we do it. Let's see if what it\n

line, we're looking for things that are between\n

But not including the to interval notation,\n

In this video, we solve linear inequalities,\n

by the conjunctions and, and or

remember when we're working with and we're\n

That is, we're looking for the overlap on\nthe number line.

In this case, the points on the number line\n

When we're working with oral statements, we're\n

other statement is true or both

this corresponds to points that are colored\n

that will actually correspond to the entire\nnumber line.

In this video, we'll solve inequalities involving\n

involving rational expressions like this one.

Let's start with a simple example. Maybe a\n

inequality, x squared is less than four, you\n

of both sides and get something like x is\n

To see why it's not correct, consider the\n

Negative 10 satisfies the inequality, x is\n

two. But it doesn't satisfy the inequality\n

10 squared is 100, which is not less than\n

same. And it doesn't work to solve a quadratic\n

both sides, you might be thinking part of\n

the negative two option, right? If we had\n

x equals two would just be one option, x equals\n

somehow, our solution to this inequality should\n

to solve an inequality involving x squares\n

equation first. But before we even do that,\n

so that my inequality has zero on the other\nside.

So for our equation, I'll subtract four from\n

Now, I'm going to actually solve the associated\n

zero, I can do this by factoring to x minus\n

I'll set my factors equal to zero, and I get\n

Now, I'm going to plot the solutions to my\n

negative two and two, those are the places\n

Since I want to find where x squared minus\n

this expression x squared minus four is positive\n

to plug in test values. So first, a plug in\n

something less than negative to say x equals\nnegative three.

If I plug in negative three into x squared\n

four, which is nine minus four, which is five,\n

the expression x squared minus four is positive.\n

the number line, my expression is going to\n

to negative without going through a place\n

x squared minus four is positive or negative\n

number line by plugging in test value similar\nway

evaluate the plug in between negative two\n

minus four, that's negative four and negative\n

minus four is negative on this whole interval.\n

10, something bigger than two, and I get 10\n

that I can tell that that's going to be a\n

Again, since I want x squared minus four to\n

on this number line where I'm getting

negatives. So I will share that in on my number\n

Because the endpoints are where my expression\n

I want it strictly less than zero

I can write my answer as an inequality, negative\n

interval notation as soft bracket negative\ntwo to soft bracket.

Our next example, we can solve similarly,\n

so that our inequality is x cubed minus 5x\n

to zero. Next, we'll solve the associated\n

down the equation. Now I'll factor out an\n

solutions to my equation are x equals 0x equals\n

I'll write the solutions to the equation on\nthe number line.

So that's negative one, zero, and six. That's\n

But I want to find where it's greater than\n

values, I can plug in, for example, x equals\n

or to this factored version. Since I only\n

it's sometimes easier to use the factored\n

But this factor, x minus six is also negative\n

Finally, x plus one, when I plug in negative\n

negative. And a negative times a negative\n

If I plug in something, between negative one\n

then I'm going to get a negative for this\n

positive for this third factor.

Negative times negative times positive gives\nme a positive

for a test value between zero and six, let's\ntry x equals one.

Now I'll get a positive for this factor a\n

positive times a negative times a positive\n

bigger than six, we could use say x equals\n

So my product will be positive.

Since I want values where my expression is\n

And the places where it's positive.

So my final answer will be close bracket negative\n

As our final example, let's consider the rational\n

by x minus one is less than or equal to zero.

Although it might be tempting to clear the\n

one, it's dangerous to do that, because x\n

it could also be a negative number. And when\n

you have to reverse the inequality. Although\n

way, by thinking of cases where x minus one\n

it's much easier just to solve the same way\n

so that we move all terms to the left and\n

true. So the next step would be to solve the\n

That is x squared plus 6x plus nine over x\n

That would be where the numerator is 0x squared\n

x plus three squared is zero, or x equals\n

have to do for rational expressions. And that's\n

not exist. That is, let's find where the denominator\n

I'll put all those numbers on the number line.\n

rational expression is equal to zero, and\n

exist, then I can start in with test values.\n

to work. If I plug those values into this\n

negative answer and a positive answer. The\n

number line where my denominators zero is\n

negative to positive by passing through a\n

exist, as well as passing by passing flew\n

Now I'm looking for where my original expression\n

I want the places on the number line where\n

So my final answer is x is less than one,\n

In this video, we solved polynomial and rational\n

think test values to make a sine chart.

The distance formula can be used to find the\n

if this first point has coordinates, x one,\n

x two, y two, then the distance between them\n

x two minus x one squared plus y two minus\ny one squared.

This formula actually comes from the Pythagorean\n

these two points as two of its vertices

then the length of this side is the difference\n

length of this vertical side is the difference\n

Now that Pythagoras theorem says that for\n

B, and hypotony is labeled C, we have that\n

Well, if we apply that to this triangle here

this hypotony News is the distance between\n

So but Tyrion theorem says, the square of\n

squared, plus the square of this side length,\n

equal the square of the hypothesis, that is\nd squared.

taking the square root of both sides of this\n

minus x one squared plus y two minus y one\n

we don't have to worry about using plus or\n

distance is always positive.

So we've derived our distance formula. Now\n

Let's find the distance between the two points,\n

this notation just means that P is the point\n

is the point with coordinates for two.

We have the distance formula

let's think of P being the point with coordinates\n

But as we'll see, it really doesn't matter\n

we get d is the square root of n x two minus\n

y two minus y one, so that's two minus five\nsquared.

Working out the arithmetic a little bit for\n

five squared, plus negative three squared.\n

are the square root of 34. Let's see what\n

point, x two y two instead, and a second.

Why'd x one y one, then we would have gotten\n

and added the difference of y's squared. So\n

That gives us the square root of negative\n

In this video, we use the distance formula\n

writing down the distance formula, sometimes\n

a minus and whether these are pluses or minuses.\n

formula comes from the Pythagorean Theorem.\n

And then a minus in here, because we're finding\n

find the lengths of the sides.

The midpoint formula helps us find the coordinates\n

as we know the coordinates of the endpoints.

So let's call the coordinates of this endpoint\n

other endpoint x two, y two

the x coordinate of the midpoint is going\n

of the endpoints. To get a number halfway\n

Similarly, the y coordinate of this midpoint\n

the y coordinates of the endpoints. So the\n

the average of those y coordinates.

So we see that the coordinates of the midpoint\n

Let's use this midpoint formula in an example

we want to find the midpoint of the segment\n

me draw the line segment between them. So\n

around here. But to find its exact coordinates,\n

and y one plus y two over two, where this\n

and the second point has coordinates x two\n

you decide is x one y one, which one is x\n

the same answer for the midpoint.

So let's see, I take my average of my x coordinates.\n

and the average of my Y coordinates, so that's\n

seven halves, as the coordinates of my midpoint.

In this video, we use the midpoint formula\n

by taking the average of the x coordinates\n

video is about graphs and equations of circles.\n

circle of radius five, centered at the point\nthree, two

will look something like this.

For any point, x, y on the circle, we know\nthat

the distance of that point xy from the center\n

distance formula, that distance of five is\n

of the x coordinates. That's x minus three

squared plus the difference of the y coordinates,\n

And if I square both sides of that equation,\n

In other words, 25 is equal to x minus three\n

square root and the squared undo each other.

A lot of times people will write the X minus\n

the left side of the equation and the 25 on\n

standard form for the equation of the circle.

The same reasoning can be generalized

Find the general equation for a circle with\n

For any point with coordinates, x, y on the\n

coordinates x, y and the point with coordinates\n

the distance is equal to r, but by the distance\n

between the x coordinates x minus h squared\n

minus k squared. squaring both sides as before,\n

canceling the square root and the squared,\n

x minus h squared plus y minus k squared equals\n

a circle with radius r, and center HK.

Notice that the coordinates h and k are subtracted\n

are in the distance formula, and the radius\n

this general formula, that makes it easy to\n

example, if we want the equation for a circle

of radius six, and center at zero, negative\nthree

then we have our equal six, and h k is our\n

we get x minus zero squared, plus y minus\n

or simplified this is x squared plus y plus\n

Suppose we're given an equation like this\n

of a circle, and if so what's the center was\nthe radius.

Well, this equation matches the form for a\n

equals r squared. If we let h, b five, Why\n

a negative six is like adding a six

five must be our r squared. So that means\n

is our radius. And our center is the point\n

which we could then graph by putting down\n

is a little bit more than two.

This equation might not look like the equation\n

To look like the equation of a circle, we're\n

form x minus h squared plus y minus k squared\nequals r squared.

First, I'd like to get rid of the coefficients\n

so I'm going to divide everything by nine.\n

8x minus two y plus four equals zero. Next,\n

x squared and the 8x. So write x squared plus\n

And I'll subtract the four over to the other\nside.

This still doesn't look much like the equation\n

to do something called completing the square,\n

divided by two and then square it. In other\n

squared, which is 16. I'm going to add 16\n

now I'm going to do the same thing to the\n

two is negative one, square that and I get\n

sides, but I'll put it near the y terms. So\n

I have to add it to the other side. On the\n

side, I can wrap up this expression x squared\n

To convince you, that's correct. If we multiply\n

x squared plus 4x plus 4x plus 16, or x squared\n

here. Similarly, we can wrap up y squared\n

y minus one squared. Again, I'll work out\n

minus y minus y plus one, or y squared minus\n

If you're wondering how I knew to use four\n

of eight, and the minus one came from taking\n

for a circle and standard form. And we can\n

negative one, and the radius, which is the\nsquare root of 13.

It might seem like magic that this trick of\n

it and adding it to both sides lets us wrap\n

perfect square. But to see why that works,\n

out x minus h squared, the thing that we're\n

x minus h squared, which is x minus h times\n

minus HX minus h x plus h squared, or x squared\n

out with this part, with some x squared term\n

decide what to add on in order to wrap it\n

on is h squared here, which comes from half\n

half of negative two H is a negative H, square\n

we do wrap it up, it's half of this coefficient\n

This trick of completing the square is really\n

in disguise, into the standard equation for\n

equation for a circle x minus h squared plus\n

r is the radius, and h k is the center.

We also showed a method of completing the\nsquare.

When you have an equation for a circle in\n

you rewrite it into the standard form.

This video is about graphs and equations of\nlines.

Here we're given the graph of a line, we want\n

for the equation of a line is y equals mx\n

B represents the y intercept, the y value\n

is equal to the rise over the run. Or sometimes\n

over the change in x values. Or in other words,\n

where x one y one and x two y two are points\non the line.

While we could use any two points on the line,\n

points where the x and y coordinates are integers,\n

grid points. So here would be one convenient\n

The coordinates of the first point are one,\n

five, negative one. Now I can find the slope\n

I go through a run of this distance, I go\n

gonna be a negative rise or a fall because\n

off squares. This is a run of 1234 squares\n

I got that answer by counting squares, but\n

Looking at the difference in my y values over\n

negative one minus two, that's from my difference\n

which is five minus one, that gives me negative\n

So my M is negative three fourths.

Now I need to figure out the value of b, my\n

the graph, it looks like approximately 2.75.\n

use a point that has integer coordinates that\n

point or that point, let's try this point.\n

mx plus b, that is y equals negative three\n

one, two, for my x and y. So that gives me\n

plus b, solving for B. Let's see that's two\n

three fourths to both sides, that's two plus\n

plus three fourths, which is 11. Four switches\n

So now I can write out my final equation for\n

plus 11 fourths by plugging in for m and b.\n

a horizontal line has slope zero. So if we\n

to be zero. In other words, it's just y equals\n

out what that that constant y value is, it\nlooks like it's

two, let's see this three, three and a half,\n

For a vertical line, like this one, it doesn't\n

to do the rise over the run

there's no run. So you'd I guess you'd be\n

But But instead, we just think of it as an\n

in this case, x equals negative two, notice\n

same x coordinate of negative two and the\n

So this is how we write the equation for a\nvertical line.

In this example, we're not shown a graph of\n

through two points. But knowing that I go\n

for the line. First, we can find the slope\n

the difference in x values. So that's negative\n

standard equation for the line, this is called\n

And we can plug in negative five thirds. And\n

still get the same final answer. So let's\n

negative five thirds times one plus b. And\n

thirds plus five thirds, which is 11 thirds.\n

method two uses a slightly different form\n

form and it goes y minus y naught is equal\n

y naught is a point on the line and again\n

same way by taking a difference in Y values\n

can simply plug in any point.

For example, the point one two will work we\n

Y not in this point slope form. That gives\nus y

Two is equal to minus five thirds x minus\none.

Notice that these two equations, while they\n

because if I distribute the negative five\n

I get the same equation as above.

So we've seen two ways of finding the equation\n

And using the point slope form.

In this video, we saw that you can find the\n

you can also find the equation for the line\n

the two points to get the slope and then plug\n

We saw two standard forms for the equation\nof a line

the slope intercept form y equals mx plus\n

And the point slope form y minus y naught\n

is the slope and x naught y naught is a point\non the line.

This video is about finding parallel and perpendicular\n

fourths, in other words, the rise

than any other line that's parallel to this\n

So that's our first fact to keep in mind,\n

on the other hand, we want to find a line\n

original slope of three fourths.

A perpendicular line, in other words, align\n

will have a slope, that's the negative reciprocal\n

slope. So we take the reciprocal of three\n

its opposite by changing it from positive\nto negative.

So I'll write this as a principle that perpendicular\n

get the hang of what it means to be an opposite\n

So here's the original slope, and this will\n

the slope of our perpendicular line. So for\n

reciprocal of two is one half opposite means\n

out to be, say, negative 1/3, the reciprocal\n

the opposite means I change it from a negative\n

One more example, if I started off with a\n

of that would be two sevenths. And I change\n

Let's use these two principles in some examples.

In our first example, we need to find the\n

slump, and go through the point negative three\n

to figure out the slope of this line. So let\n

the the slope intercept form. So I'll start\n

I'm going to solve for y to put it in this\n

y equals 4x minus six and then divide by three\n

is divided all my turns by three, I can simplify\n

two. Now I can read off the slope of my original\n

slope of my new line, my parallel line will\n

the equation y equals four thirds ax plus\n

be negative two like it was for the

first line, it'll be something else determined\n

negative three to, to figure out what b is,\n

negative three for x, and two for y. So that\n

three plus b, and I'll solve for b. So let's\n

this is negative 12 thirds plus b, or in other\n

means that B is going to be six, and by my\n

Next, let's find the equation of a line that's\n

go through a given point. Again, in order\n

of this given line is. So I'll rewrite it,\n

three, y equals four. And then I can put it\n

for y. So three, y is negative 6x plus four,\n

negative six thirds x plus four thirds. So\n

original slope of my original line is negative\n

the opposite reciprocal, so I take the reciprocal\n

and I change the sign so that gives me one\n

Now, my new line, I know is going to be y\n

can plug in my point on my new line, so one\n

I get one equals two plus b, So b is negative\n

These next two examples are a little bit different,\n

to a completely horizontal line, let me draw\n

is always three, which means that my line\n

at height y equals three, if I want something\n

line. Since it goes through the point negative\n

it's gonna have always have a y coordinate\n

point it goes through, so my answer will just\nbe y equals one.

In the next example, we want a line that's\n

horizontal line y equals four, but perpendicular\n

I need a vertical line that goes through the\npoint three, four.

Okay, and so I'm going to draw a vertical\n

x equals something for the equation. And to\n

the x coordinate of the point I'm going through\n

is three and all the points on this, this\n

three so my answer is x equals three.

In this video will use the fact that parallel\n

lines have opposite reciprocal slopes, to\n

This video introduces functions and their\ndomains and ranges.

A function is a correspondence between input\n

usually the y values, that sends each input\n

a function is thought of as a rule or machine

in which you can feed in x values as input\n

example of a function might be the biological\n

person and it gives us output their biological\nmother.

This function satisfies the condition that\n

get sent to exactly one output per

Because if you take any person, they just\n

give you a function. But if I change things\n

which sends to each person, their mother,\n

there are some people who have more than one\n

mother and adopted mother, or a mother and\n

So since there's, there's at least some people\n

get like more than one possible output that\n

would not be a function. Now, most of the\n

with equations, not in terms of mothers. So\n

This can also be written as f of x equals\n

notation that stands for the output value\nof y.

Notice that this notation is not representing\n

x, instead, we're going to be putting in a\n

of f of x or y. For example, if we want to\n

input for x either in this equation, or in\n

plus one, f of two is going to equal five.\n

for x, so that's gonna be five squared plus\n

a function on a more complicated expression\n

the functions value on any expression, it's\n

that whole expression for x.

So f of a plus three is going to be the quantity\n

we could rewrite that as a squared plus six\n

When evaluating a function on a complex expression,\n

you plug in for x. That way, you evaluate\n

to write f of a plus three equals a plus three\n

because that would imply that we were just\n

Sometimes a function is described with a graph\n

graph is supposed to represent the function\n

functions. For example, the graph of a circle\n

the graph of a circle violates the vertical\n

intersect the graph in more than one point.

But our graph, it left satisfies the vertical\n

graph, and at most one point, that means is\n

at most one y value that corresponds to it.\n

an x value. And we'll use the graph to find\n

on the x axis, and find the point on the graph\n

I can look at the y value of that point looks\n

to three. If I try to do the same thing to\n

is an x value, I look for it on the x axis,\n

Therefore, g of five is undefined, or we can\n

x values and y values make sense for a function\n

The domain of a function is all possible x\n

The range is the y values that make sense\nfor the function.

In this example, we saw that the x value of\n

this function. So the x value of five is not\n

x values in the domain, we have to look at\n

the graph. One way to do that is to take the\n

onto the x axis and see what x values are\n

starting at negative eight, and continuing\n

is the x's between negative eight, and four,\n

this in interval notation as negative eight\n

To find the range of the function, we look\n

By taking the shadow or projection of the\ngraph onto the y axis

we seem to be hitting our Y values from negative\n

So our range is wise between negative five\n

five, three with square brackets. If we meet\n

instead of a graph, one way to find the domain\n

often possible to find the domain at least\n

We think about what x values that makes sense\n

need to be excluded, because they make the\n

Specifically, to find the domain of a function,\n

denominator zero. Since we can't divide by\nzero

we also need to exclude x values that make\n

Since we can't take the square root of a negative\n

that make the expression inside any even root\n

of a negative number, even though we can take\n

number. Later, when we look at logarithmic\n

that we have to make. But for now, these two\n

see. So let's apply them to a couple examples.

For the function in part A, we don't have\n

So we need to exclude x values that make the\n

x squared minus 4x plus three to not be equal\nto zero.

If we solve x squared minus 4x plus three\n

And that gives us x equals three or x equals\n

All other x values should be fine. So if I\n

three and just dig out a hole at both of those.\n

the number line. In interval notation, this\n

infinity to one, together with everything\n

In the second example, we don't have any denominator\n

sign. So we need to exclude any x values that\n

words, we can include all x values for which\n

zero. Solving that inequality gives us three\n

is less than or equal to three halves

I can draw this on the number line

or write it in interval notation.

Notice that three halves is included, and\n

be zero, I can take the square root of zero,\n

Finally, let's look at a more complicated\n

and denominator. Now there are two things\nI need to worry about.

to not be equal to zero, and I need the stuff\n

than or equal to zero. from our earlier work,\n

not equal to three, and x is not equal to\n

is less than or equal to three halves. Let's\n

x is not equal to three and x is not equal\n

two dug out points. And the other condition\n

we can have three halves and everything

To the left of it. Now to be in our domain\n

both of these conditions to be true. So I'm\n

N, that means we're looking for a numbers\n

and blue. So I'll draw that above in purple.\n

one, I have to dig out one because one was\n

can continue for all the things that are colored\n

domain is going to be, let's see negative\n

with one, but not including it to three halves,\n

in this video, we talked about functions,\n

and ranges. This video gives the graphs of\n

toolkit functions. The first function is the\n

on the graph of this function. If x is zero,\n

is always equal to x doesn't have to just\n

and we'll connecting the dots, we get a straight\n

Let's look at the graph of y equals x squared.\n

the origin again, if x is one, y is one, and\n

the x value of two gives a y value of four\n

value of four also connecting the dots, we\nget a parabola.

That is this, this function is an even function.

That means it has mirror symmetry across the\n

like the mirror image of the right side. That\n

number, like two, you get the exact same y\n

The next function y equals x cubed. I'll call\n

x is zero, y is zero. When x is one, y is\n

one, two goes with the point eight way up\n

to give us negative eight. If I connect the\n

This function is what's called an odd function,\n

occur around the origin. If I rotate this\n

turn the paper upside down, I'll get exactly\n

this odd symmetry is because when I cube a\n

n cube the corresponding negative number to\n

gives us exactly the negative of the the y\n

Let's look at the next example. Y equals the\nsquare root of x.

Notice that the domain of this function is\n

because we can't take the square root of a\n

X is zero gives y is 0x is one square root\n

and connecting the dots, I get a function\nthat looks like this.

The absolute value function is next. Again,\n

with y equals 0x is one gives us one, the\n

the graph, and the absolute value of negative\n

graph. It also has even or a mirror symmetry.

Y equals two to the x is what's known as an\n

x is in the exponent. If I plot a few points

two to the one is two, two squared is four,\n

these on my graph, you know, let me fill in\n

is eight. That's way up here and negative\ntwo gives

Maybe 1/4 1/8 connecting the dots, I get something\n

You might have heard the expression exponential\n

growth, this is function is represents exponential\n

time we increase the x coordinate by one,\n

We could also look at a function like y equals\n

where E is just a number about 2.7. These\n

bigger bass makes us rise a little more steeply.

Now let's look at the function y equals one\n

I can plug in x equals one half, one over\none half is to

whenever one is one, and one or two is one\n

in the first quadrant, but I haven't looked\n

whenever negative one is negative one, whenever\n

similar looking piece in the third quadrant.

This is an example of a hyperbola.

And it's also an odd function, because it\n

I turn the page upside down, it'll look exactly\n

Again, it's not defined when x is zero, but\n

see one over one half squared is one over\n1/4, which is four.

And one over one squared, one over two squared\n

previous function is just a little bit more\n

a little more dramatically. But for negative\n

goes on. For example, one over negative two\n

fourth. So I can plot that point there, and\n

one. So my curve for negative values of x\n

third quadrant. This is an example of an even\nfunction

because it has perfect mirror symmetry across\nthe y axis.

These are the toolkit functions, and I recommend\n

That way, you can draw at least a rough sketch\n

That's all for the graphs of the toolkit functions.

If we change the equation of a function, then\n

This video gives some rules and examples for\n

out of this video, it's helpful if you're\n

functions, I call them toolkit functions like\n

y equals the absolute value of x and so on.\n

I encourage you to watch my video called toolkit\n

I want to start by reviewing function notation.

If g of x represents the function, the square\n

in terms of square roots. For example, g of\n

g of quantity x minus two means we plug in\n

would be the same thing as the square root\n

In this second example, I say that we're subtracting\n

we're subtracting two before we apply the\n

example, I say that the minus two is on the\n

root function first and then subtracting two.\n

by three on the inside of the function. To\n

plug in the entire 3x for x and the square\n

In the next example, we're multiplying by\n

is just three times the square root of x.\n

Now, this might look a little odd because\n

a negative number. But remember that if x\n

x will be negative negative two or positive\n

root of a positive number in that case, let\n

of these are outside of my function.

In this next set of examples, we're using\n

this time, we're starting with an expression\n

it in terms of g of x. So the first example,\n

because I'm taking the square root of x first,\n

In the second example, I'm taking x and adding\n

whole thing. Since I'm adding the 12 to x\n

So I write that as g of the quantity x plus\n12.

Remember that this notation means I plug in\n

sign, which gives me exactly square root of\n

the square root first and then multiplying\n

outside my function, I can rewrite this as\n

example, I take x multiplied by a fourth and\n

the same thing as g of 1/4 X, my 1/4 X is\n

it's inside the parentheses when I use function\nnotation.

Let's graph the square root of x and two transformations\n

y equals the square root of x goes to the\n

root of four is two, it looks something like\nthis.

In order to graph y equals the square root\n

is on the outside of the function, that means\n

and then subtract two. So for example, if\n

the square root of zero, that's zero, then\n

an x value of one, which under the square\n

a y value that's decreased by two, one minus\n

of four, which under the square root function\n

two minus two or zero, its y value is also\n

zero goes with negative two, one goes with\n

Because I subtracted two on the outside of\n

two, which brought my graph down by two units.\n

of quantity x minus two. Now we're subtracting\n

we subtract two from x first and then take\n

y value of zero as we had in our blue graph,\n

need our original x to be two.

In order to get the y value of one that we\n

the square root of one, so we need x minus\n

start with an x value of three.

And in order to reproduce our y value of two\n

root of x minus two to be two, which means\n

root of four, which means our x minus two\n

If I plot my x values, with my corresponding\n

get the following graph. Notice that the graph\n

To me, moving down by two units, makes sense\n

y's by two units, but the minus two on the\n

I expect, I might expected to to move the\n

be going down by two units, but instead, it\n

because the x units have to go up by two units\n

I then subtract two units again, the observations\n

hold in general, according to the following\n

In our example, y equals a squared of x minus\n

a result in vertical motions, like we saw,\n

So subtracting two was just down by two. If\n

numbers on the inside of the function. That's\n

of quantity x minus two, those affect the\n

these motions go in the opposite direction\n

two on the inside actually shifted our graph\n

the inside, that would actually shift our\ngraph to the left.

Adding results in a shift those are called\n

like y equals three times the square root\n

In other words, if I start with the square\nroot of x

and then when I graph y equals three times\n

vertically by a factor of three.

if I want to graph y equals 1/3, times the\n

Finally, a negative sign results in reflection.\n

y equals the square root of x, and then when\n

x, that's going to do a reflection in the\n

is on the inside of the square root sign.

A reflection in the horizontal direction means\n

If instead, I want to graph y equals negative\n

outside means a vertical reflection, a reflection\n

Pause the video for a moment and see if you\n

In the first example, we're subtracting four\n

subtracting means a translation or shift.\n

affects the y value, so that's moving us vertically.\n

root of graph and move it down by four units,\n

In the next example, we're adding 12 on the\n

we're moving horizontally. And so since we\n

going to go to the left by 12 units, that's\n

And the next example, we're multiplying by\n

on the outside of our function outside our\n

So in multiplication means we're stretching\n

we reflect in the vertical direction

here's stretching by a factor of three vertically,\n

minus sign reflects in the vertical direction.

Finally, in this last example, we're multiplying\n

we know that multiplication means stretch\n

it's a horizontal motion, and it does the\n

shrinking by a factor of 1/4, horizontally,\n

a factor of four horizontally.

that'll look something like this. Notice that\n

looks kind of like shrinking vertically by\na factor of one half.

And that's actually borne out by the algebra,\n

thing as the square root of 1/4 times the\n

as one half times the square root of x. And\n

shrink by a factor of one half is the same\n

at least for this function, the square root\nfunction.

This video gives some rules for transformations\n

on the outside correspond to changes in the\n

numbers on the inside of the function, affect\n

corresponds to translations or shifts.

multiplying and dividing by numbers corresponds\n

and putting in a negative sign.

horizontal reflection, if the negative sign\n

if the negative sign is on the outside

knowing these basic rules about transformations\n

much more complicated functions, like y equals\n

by simply considering the transformations,\none at a time.

a quadratic function is a function that can\n

plus bx plus c, where a, b and c are real\n

we require that A is not equal to zero is\n

f of x is equal to b x plus c, which is called\n

a is not zero, we make sure there's really\n

Please pause the video for a moment and decide\n

The first function can definitely be written\n

bx plus c. fact it's already written in that\n

The second equation is also a quadratic function,\n

one times x squared plus zero times x plus\nzero.

So it is in the right form, where A is one,\n

It's perfectly fine for the coefficient of\n

function, we just need the coefficient of\n

The third equation is not a quadratic function.\n

The fourth function might not look like a\n

expanding out the X minus three squared, let's\n

x minus three times x minus three plus four.\n

3x, plus nine plus four, continuing, I get\n2x squared

minus 12 access, plus 18 plus four, in other\n

22. So in fact, our function can be written\n

A function that is already written in the\n

said to be in standard form.

So our first example g of x is in standard\nform

a function that's written in the format of\n

equals a times x minus h squared plus k for\n

I'll talk more about standard form and vertex\n

In this video, we identified some quadratic\n

This video is about graphing quadratic functions.\n

in standard form, like this, or sometimes\n

graph that looks like a parabola.

This video will show how to tell whether the\n

its x intercepts, and how to find its vertex.

The bare bones basic quadratic function is\n

since f of zero is zero squared, which is\n

The vertex of a parabola is its lowest point\n

point if it's pointing downwards. So in this\n

The x intercepts are where the graph crosses\n

In this function, y equals zero means that\n

x is zero. So the x intercept, there's only\n

The second function, y equals negative 3x\n

functions value when x is zero, is y equals\n

That's because thinking about transformations\n

reflects the function vertically over the\n

upwards, reflecting the point downwards.

The number three on the outside stretches\n

So it makes it kind of long and skinny like\nthis.

In general, a negative coefficient to the\n

down. Whereas a positive coefficient, like\n

Alright, that roll over here.

So if a is bigger than zero, the parabola\nopens up.

And if the value of the coefficient a is less\n

In this second example, we can see again that\n

Let's look at this third example.

If we multiplied our expression out, we'd\n

be to a positive number. So that means our\n

But the vertex of this parabola will no longer\n

parabolas vertex by thinking about transformations\nof functions.

Our function is related to the function y\n

by three and up by four. Since y equals 2x\n

whole parabola including the vertex, right\n

end up at the point three, four.

So a parabola will look something like this.

Notice how easy it was to just read off the\n

in this form. In fact, any parabola any quadratic\n

h squared plus k has a vertex at h k. By the\n

with a vertex at the origin to the right by\nH, and by K.

That's why this form of a quadratic function\n

Notice that this parabola has no x intercepts\n

For our final function, we have g of x equals\n

we know the graph of this function will be\n

To find the x intercepts, we can set y equals\n

graph crosses the x axis, and that's where\nthe y value is zero.

So zero equals 5x squared plus 10x plus three,\n

solve that. So x is negative 10 plus or minus\n

five times three, all over two times five.\n

or minus the square root of 40 over 10

which simplifies further to x equals negative\n

of 40 over 10, which is negative one plus\n

negative one plus or minus square root of\n10 over five.

Since the square root of 10 is just a little\n

about negative two fifths and negative eight\n

our parabola is going to look something like\n

the x axis at y equals three, that's because\n

we get y equal three, so the y intercept is\n

Since this function is written in standard\nform

On the Y equals a x plus a squared plus bx\n

just read off the vertex like we couldn't\n

formula, which says that whenever you have\n

form, the vertex has an x coordinate

of negative B over two A. So in this case,\n

two times five or negative one, which is kind\n

the y coordinate of that vertex, I can just\n

x, which gives me at y equals five times negative\n

So I think I better redraw my graph a little\n

of negative two where it's supposed to be.

Let's summarize the steps we use to graph\n

graph of a quadratic function has the shape\nof a parabola.

The parabola opens up, if the coefficient\n

than zero and down if a is less than zero.\n

zero, or in other words, f of x equal to zero\n

either read it off as h k, if our function\nis in vertex form

or we can use the vertex formula

of the vertex to be negative B over to a if\n

To find the y coordinate of the vertex in\n

Finally, we can always find additional points\n

In this video, we learned some tricks for\n

we saw that the vertex can be read off as\n

form, and the x coordinate of the vertex can\nbe calculated

as negative B over two A, if our function\n

why this vertex formula works, please see\nthe my other video.

a quadratic and standard form looks like y\n

b and c are real numbers, and a is not zero.\n

like y equals a times x minus h squared plus\n

is not zero. When a functions in vertex form,\n

This video explains how to get from vertex\n

Let's start by converting this quadratic function\n

pretty straightforward, we just have to distribute\nout.

So if I multiply out the X minus three squared

I get minus four times x squared minus 6x\n

four, I get negative 4x squared plus 24x minus\n

squared plus 24x minus 35. And I have my quadratic\n

Now let's go the other direction and convert\n

form into vertex form. That is, we want to\n

x minus h squared plus k, where the vertex\n

The vertex formula says that the x coordinate\n

over to a, where A is the coefficient of x\n

in this case, we get an x coordinate of negative\n

To find the y coordinate of the vertex, we\n

for a g of x. So that's g of negative two,\n

eight times negative two plus six. And that\n

So the vertex for our quadratic function has\n

if I want to write g of x in vertex form,\n

two squared plus minus two. That's because\n

simplifies to g of x equals a times x plus\n

need to figure out what this leading coefficient\n

this out, then the coefficient of x squared\n

of x squared here, which is a has to be the\n

which we conveniently also called a, in other\n

I'm going to write that as g of x equals two\n

twos in this problem. And that's our quadratic\n

my answer, of course, I could just distribute\n

4x, plus two, minus two, in other words, 2x\n

to exactly what I started with. This video\n

form by distributing out and how to get from\n

vertex using the vertex formula.

Suppose you have a quadratic function in the\n

you want to find where the vertex is, when\nyou graph it.

The vertex formula says that the x coordinate\n

this video gives a justification for where\n

Let's start with a specific example. Suppose\n

vertex for this quadratic function. To find\n

and solve for x. So that's zero equals 3x\n

x, I use the quadratic formula. So x is going\n

root of seven squared minus four times three\n

That simplifies to negative seven plus or\n

I could also write this as negative seven,\n

or x equals negative seven, six minus the\n

root of 109 is just a little bit bigger than\n

six plus 10, six, and negative seven, six\nminus 10, six.

So pretty close to, I guess about what half\n

three over here, I'm just going to estimate\n

Since the leading coefficient three is positive,\n

up and the intercepts are somewhere around\n

Now the vertex is going to be somewhere in\n

to be by symmetry, it'll be exactly halfway\n

intercepts are negative seven, six plus and\n

halfway in between those is going to be exactly\n

right, because on the one hand, I have negative\n

hand, I have negative seven six minus that\nsame thing.

So negative seven, six will be exactly in\n

will be at negative seven sex. Notice that\n

More generally, if I want to find the x intercepts\n

and solve for x using the quadratic formula,\n

b squared minus four AC Oliver to a, b, x\n

the x coordinate of the vertex, which is exactly\n

be at negative B over two A. That's where\n

And it turns out that this formula works even\n

the quadratic formula gives us no solutions.\n

And that's the justification of the vertex\nformula.

This video is about polynomials and their\ngraphs.

We call that a polynomial is a function like\n

His terms are numbers times powers of x. I'll\n

a polynomial is the largest exponent. For\n

the leading term is the term with the largest\n

term is 5x to the fourth, it's conventional\n

of powers of x. So the leading term is first.\n

first term. If I wrote the same polynomial\n

cubed plus 5x. Fourth, the leading term would\n

The leading coefficient is the number in the\n

Finally, the constant term is the term with\n

please pause the video for a moment and take\n

figure out what's the degree the leading term,\n

The degree is again, four, since that's the\n

is negative 7x. to the fourth, the leading\n

In the graph of the polynomial Shown here\n

points, because the polynomial turns around\n

those same points can also be called local\n

maximum and minimum points. For this polynomial,\n

Let's compare the degree and the number of\n

For the first one, the degree is to and there's\n

For the second example, that agree, is three,\n

And for this last example, the degree is four,\n

For this first example, and the next two,\n

less than the degree. So you might conjecture\n

is not always true. In this last example,\n

In fact, it turns out that while the number\n

minus one, it is always less than or equal\n

Remember, when you're sketching graphs are\n

The end behavior of a function is how the\n

and bigger heads towards infinity, or x gets\n

In this first example, the graph of the function\n

goes towards negative infinity. I can draw\n

Down on either side, or I can say in words,\n

and falling also as we had right.

In the second example, the graph rises to\n

example, the graph falls to the left, but\n

it rises to the left and falls to the right.\n

might notice there's a relationship between\n

of the polynomials and the end behavior. Specifically,\n

by whether the degree is even or odd. And\n

When the degree is even, and the leading coefficient\n

leading coefficient is one, we have this sorts\n

When the degree is even at the leading coefficient\n

the end behavior that's falling on both sides

when the degree is odd, and the leading coefficient\n

the degree three and the leading coefficient\n

And finally, when the degree is odd, and the\n

example, we have this sort of NBA havior.

I like to remember this chart just by thinking\n

y equals negative x squared, y equals x cubed,\n

by heart what those four examples look like

then I just have to remember that any polynomial\n

has the same end behavior as x squared.

And similarly, any polynomial with even degree\n

end behavior as negative x squared. And similar\n

We can use facts about turning points and\n

of a polynomial just by looking at this graph.

In this example, because of the end behavior,\n

we know that the leading coefficient

And finally, since there are 1234, turning\n

That's because the number of turning points\n

one. And in this case, the number of turning\n

And so solving that inequality, we get the\n

Put in some of that information together,\n

five, or seven, or nine, or any odd number\n

be for example, three or six. Because even\n

This video gave a lot of definitions, including\n

We saw that knowing the degree and the leading\n

about the number of turning points and the\n

This video is about exponential functions\nand their graphs.

an exponential function is a function that\n

times b to the x, where a and b are any real\n

It's important to notice that for an exponential\n

This is different from many other functions\n

function like f of x equals 3x squared has\n

For exponential functions, f of x equals a\n

to zero, because otherwise, we would have\n

Which just means that f of x equals zero.\n

an exponential function. Because f of x is\nalways equal to zero

in an exponential function, but we require\n

for example, if b is equal to negative one,\n

one to the x. Now, this would make sense for\n

something like f of one half, with our Bs,\n

as a times the square root of negative one,\n

we'd get the same problem for other values\n

if we tried b equals zero, we'd get a kind\n

which again is always zero. So that wouldn't\n

use any negative basis, and we can't use zero\n

The number A in the expression f of x equals\n

And the number B is called the base.

The phrase initial value comes from the fact\n

times b to the zero, well, anything to the\n

other words, f of zero equals a. So if we\nthink of starting out

when x equals zero, we get the y value

of a, that's why it's called the initial value.

Let's start out with this example, where y\n

and we've set a equal to one and B equals\n

when x is zero, is going to be one.

If I change my a value, my initial value,\n

If I make the value of a go to zero, and then\nnegative

then my initial value becomes negative, and\n

back to an a value of say one, and see what\n

value the basis two, if I increase B, my y\n

steeper and steeper. If I put B back down\n

exactly one, my graph is just a constant.

As B gets into fractional territory, point\n

other way, it's decreasing now instead of\n

still hasn't changed, I can get it more and\n

away from one of course, when B goes to negative\n

So a changes the y intercept, and B changes\n

whether it's increasing for B values bigger\n

we'll summarize all these observations on\nthe next slide.

We've seen that for an exponential function,\n

or number a gives the y intercept, the parameter\n

decreasing. Specifically, if b is greater\n

And if b is less than one, the graph is decreasing.

The closer B is to the number one, the flatter\nthe graph.

So for example, if I were to graph y equals\n

four to the x, they would both be decreasing\n

less than one. But point two five is farther\n

one. So point four is going to be flatter.\n

So in this picture,\nThis red graph would correspond to point two

five to the x, and the blue graph would correspond\n

functions, whether the graphs are decreasing\n

asymptote along the x axis. In other words,\n

always from negative infinity to infinity,\n

because the range is always positive y values.\n

zero, if a is less than zero, then our graph\n

same, but our range becomes negative infinity\n

is f of x equals e to the x. This function\n

x of x. The number E is Oilers number as approximately\n

to calculus and to some compound interest\n

functions, functions of the form a times b\n

we saw that they all have the same general\n

like this, unless a is negative, in which\n

a horizontal asymptote at y equals zero, the\n

functions to model real world examples. Let's\n

salary is $40,000. With a guaranteed annual\n

salary be after one year, two years, five\n

me chart out the information. The left column\n

And the right column will be your salary.\n

hired, your salary will be $40,000. After\n

your salary will be the original 40,000 plus\n

of this first number as one at times 40,000.\n

terms, to get 40,000 times one plus 0.03.\n

is your original salary multiplied by a growth\n

a 3% raise from your previous year salary,\n

3% of that, again, I can think of the first\n

I can factor out the common factor of 40,000\n

times 1.03 times one plus 0.03. Let me rewrite\n

times 1.3 squared. We can think of this as\n

growth factor of 1.03. After three years,\n

new salary is your previous year salary times\n

as 40,000 times 1.03 cubed. And in general,\n

years, your salary should be 40,000 times\n

salary after two years is your original salary\n

to the t power, let me write this as a formula\n

to 40,000 times 1.03 to the T. This is an\n

the form a times b to the T, where your initial\n

Notice that your base is the amount that your\n

this formula, we can easily figure out what\n

years by plugging in five for T. I worked\n

to the nearest cent. exponential functions\n

The United Nations estimated that the world\n

at a rate of 1.1% per year. Assuming that\n

to stay the same, we'll write an equation\n

2010 1.1%, written as a decimal is 0.011.\n

that after zero years since 2010, we have\n

one year, we'll take that 6.7 9 billion and\n

This works out to 6.79 times one plus point\n

initial population of 6.7 9 billion, and our\n

population got multiplied by in one year.\n

years, our population becomes 6.79 times 1.011\n

twice, and after two years, it'll be 6.79\n

that models population is going to be 6.79\n

in years, since 2010. Just for fun, I'll plug\n

So that's the year 2050. And I get 6.79 times\n

10 point 5 billion. That's the prediction\n

The previous two examples were examples of\n

example of exponential decay. The drugs Seroquel\n

at a rate of 11% per hour. If 400 milligrams\n

hours later, I'll chart out my information\n

since the dose was given, and the right column\n

still on the body. zero hours after the dose\n

in the body. One hour later, we have the formula\n

point one one times 400. If I factor out the\n

0.11 or 400 times point eight nine. The 400\n

nine I'll call the growth factor, even though\n

growing. So really it's kind of a shrink factor.\n

I'll have 400 times 0.89 my previous amount,\n

nine, so that's going to be 400 times 0.89\n

have 400 multiplied by this growth or shrinkage\n

t power. Since each hour, the amount of Seroquel\n

number less than one. All right, my exponential\n

0.89 to the T, where f of t represents the\n

And t represents the number of hours since\n

in the body after 24 hours, I just plug in\n

I hope you notice the common form for the\n

examples. The functions are always in the\n

a represented the initial amount, and B represented\n

B, we started with the percent increase or\n

we either added or subtracted it from one,\n

or decreasing. Let me show you that as a couple\n

increase of 3% on the race as a decimal, I'll\n

get the growth factor, we added that to one\n

we had a 1.1% increase, we wrote this as point\n

drug example, we had a decrease of 11%. We\n

the point one one from one to get 0.89. In\n

B as one plus the percent change written as\n

percent change, negative when the quantity\n

is increasing. Since here one plus negative\npoint 11

gives us the correct growth factor of point\n

check. Remember that if your quantity is increasing,\n

if the quantity is decreasing, then B should\nbe less than one.

exponential functions can also be used to\n

interest, as we'll see in another video. This\n

An antique car is worth $50,000 now, and its\n

an equation to model its value x years from\n

plus 0.07 times the 50,000. That's because\n

times 50,000. this can be written as 50,000\n

adding 7% to the original value is the same\n

point oh seven or by 1.07. After two years,\n

squared, or 50,000 times 1.07 squared. That's\n

again by 1.7. In general, after x years We\n

times 1.07 to the x. That's because the original\n

one time for each year. If we dissect this\n

comes from the original value of the car.\n

factor comes from one plus point 707. The\n

written as a decimal. So the form of this\n

x equals a times b to the x, where A is the\n

But we could also write this as a times one\n

R is the percent increase written as a decimal.\n

example, here, my Toyota Prius is worth only\n

5% each year. So after one year, its value\n

3000 times one minus 0.05. I can also write\n

by 5% is like multiplying the value by one\n

After two years, the value will be multiplied\n

be 3000 times point nine, five squared. And\n

point nine five to the x. So my equation for\n

the x. This is again, an equation of the form\n

a is 3000, the initial value, and B is point\n

factor, even though we're actually declining\n

point nine five came from, it came from taking\n

5%, decrease in value, so I can again, write\n

time times one minus r to the x, where R is\n

written as a decimal. Please take a moment\n

They say that when you have an exponential\n

If it's written in this form, B is your growth\n

one minus r, where r is the percent decrease,\n

In this example, we're given a function f\n

petri dish x hours after 12 o'clock noon,\n

at noon, and by what percent, the number of\n

see from the equation that the number of bacteria\n

the base of the exponential function 1.45\n

f of x equals 12 times 1.45 to the X has the\n

of it as a times one plus r to the x. Here\n

this familiar form, we can recognize that\n

be 12 12,000. Since those are our units and\n

the number of bacteria is multiplied by each\n

the rate of increase, in other words, a 45%\n

questions are 12,045%. In this example, the\n

exponential function, where x is the number\n

of salamanders is decreasing, because the\n

eight is less than one. So if we recognize\n

b to the x, or we can think of this as a times\n

value, and r is our percent decrease written\nas a decimal.

Our initial value is 3000. So that's the number\n

growth factor B is 0.78. But if I write that\n

minus 0.78, or 0.22. In other words, our population\n

we saw that exponential functions can be written\n

x, where A is the initial value. And B is\n

be written in the form a times one plus r\n

as a times one minus r to the x when the amount\n

increase or the percent decrease written as\n

of 0.15 and a growth factor B of 1.15. Whereas\n

one, two, and a B value of one minus point\n

help us quickly interpret exponential functions.\n

of 100 and a 15% increase. And here, we have\n

exponential functions can be used to model\n

Suppose you invest $200 in a bank account\n

make no deposits or withdrawals, how much\n

because 3% of the money that's in the bank\n

bank gets multiplied by 1.03 each year. So\n

1.3, after two years 200 times one point O\n

one point O three to the t power. So the function\n

P of t is given by 200 times 1.03 to the T.\n

at an annual interest rate of our for t years.\n

here r needs to be written as a decimal, so\n

rate. Going back to our specific example,\n

to be P of 10 which is 200 times 1.03 to the\n

cent. In this problem, we've assumed that\n

in the next few examples, we'll see what happens\n

twice a year, or every month. For example,\n

4.5% annual interest compounded semi annually,\n

A 4.5% annual interest rate compounded two\n

4.5 over 2% interest, every time the interest\n

2.25% interest every half a year. Note that\n

So every time we earn interest, our money\n

a chart of what happens. After zero years,\n

original $300. for half a year, that's one\n

time, so we multiply the 300 by 1.02 to five,\n

money earns interest two times. So we multiply\n

after 1.5 years, that's three half years,\n

after two years or four half years, we have\n

after t years, which is to T half years, our\n

to the two t power. Because we've compounded\n

amount of money is P of t equals 300 times\n

t is the number of years. To finish the problem,\n

is 300 times 1.02 to five to the two times\n

to the nearest cent. In this next example,\n

in annual interest rate of 6% compounded monthly.\n

different, they're mathematically the same.\n

investing money in you and getting interest\n

out with the same kind of math 6% annual interest\n

12 times a year. So each time you compound\n

12% interest. That's point 5% interest. And\n

again, what happens. Time is zero, of course,\n

After one year, that's 12 months, your loan\n

gets multiplied by 1.05 to the 12th.

After two years, that's 24 months, it's had\n

multiplied by 1.05 to the 24th power. Similarly,\n

amount will be 1200 times 1.05 to the 36th\n

12 t months. So the interest will be compounded\n

To the 12 t power. This gives us the general\n

1200 times 1.05 to the 12 T, where T is the\n

years, we'll have to pay back a total of 1200\n

which works out to $1,436.02 to the nearest\n

pattern. If A is the initial amount of the\n

interest rate, compounded n times per year,\n

going to be a times one plus r over n to the\n

did in this problem. First, we took the interest\n

number of compounding periods each year 12.\n

it. That's where we got the 1.05 from. We\n

but to 12 times the number of years. That's\n

times the number of years. And we multiplied\n

was 1200. This formula for compound interest\n

to be able to reason your way through it,\n

type of compound interest. And that's interest\n

continuous compounding as the limit of compounding\n

100 times a year 1000 times a year a million\n

compounding. The formula for continuous compounding\n

p of t is the amount of money, t is the time\n

And R is the annual interest rate written\n

the 2.5% annual interest rate. He represents\n

is about 2.718. So in this problem, we have\n

after five years, we'll have P of five, which\n

which works out to $4,532.59 to the nearest\n

interest rate written as a decimal, that is\n

of years and a represents the initial amount\n

compounded once a year. Our formula is P of\n

interest compounded n times per year. Our\n

n to the n T. And for compound interest compounded\n

In this video, we looked at three kinds of\n

interest, interest compounded and times per\n

This video introduces logarithms. logarithms\n

log base a of B equals c means that a to the\n

B is the exponent that you raise a to to get\n

logarithm. It's also called the base when\n

students find it helpful to remember this\n

a to the C equals b, by drawing arrows

Other students like to think of it in terms\n

What power do you raise a to in order to get\n

of eight is three, because two to the three\n

y is asking you the question, What power do\n

log base two of 16 is four, because it's asking\n

And the answer is four. Please pause the video\n

base two of two is asking, What power do you\n

one. Two to the one equals two. log base two\n

you one half? Well, to get one half, you need\n

would be two to the negative one. So the answer\n

what power do we raise to two in order to\n

we have to raise two to the negative three\n

is negative three. And that's our answer to\n

of one is asking to what power equals one.\n

us one, so this log expression evaluates to\n

and zero answers for our logarithm expressions.\n

these logs evaluate to. to work out log base\n

10 to the sixth power. Now we're asking the\n

get a million? So that is what power do we\n

course, the answer is going to be six. Similarly,\n

this log expression is the same thing as asking,\n

three? Well, what power do you have to raise\n

the answer is negative three. Log base 10\n

10 to to get zero. If you think about it,\n

get zero. Raising 10 to a positive exponent\n

10 to a negative exponent is like one over\n

but they're still positive numbers, we're\n

10 to the zero power, we'll just get one.\n

base 10 of zero does not exist. If you try\n

button, you'll get an error message. Same\n

100. We're asking 10 to what power equals\n

will work. And more generally, it's possible\n

than zero, but not for numbers that are less\n

domain of the function log base a of x, no\n

is going to be all positive numbers. A few\n

called natural log, and it means the log base\n

about 2.718. When you see log of x with no\n

base 10 of x And it's called the common log.\n

natural log, and for common log. Let's practice\n

base three of one nine is negative two, can\n

Log of 13 is shorthand for log base 10 of\n

1.11394 equals 13. Finally, in this last expression,\n

can rewrite this equation as log base e of\n

means the same thing as e to the negative\n

let's go the opposite direction. We'll start\n

as logs. Remember that log base a of B equals\n

b, the base stays the same in both expressions.\n

the exponential equation, that's going to\n

just have to figure out what's in the argument\n

of the equal sign. Remember that the answer\n

goes in this box should be my exponent for\n

And I'll put the 9.78 as the argument of my\n

9.78 equals view means the same thing as three\n

started with. In the second example, the base\n

of my log is going to be the answer to my\n

3x plus seven. And the other expression, the\n

Let me check, log base e of four minus y equals\n

equals four minus y, which is just what I\n

e as natural log. This video introduced the\n

of B equal c means the same thing as a to\n

you the question, What power exponent Do you\n

we'll work out the graph, so some log functions\n

first example, let's graph a log function\n

we're working with is y equals log base two\n

Since we're working this out by hand, I want\n

log base two of x. So I'll start out with\n

one is zero, log base anything of one is 02\n

log base two of two, that's asking, What power\n

is one. Power other powers of two are easy\n

of four that saying what power do I raise\n

Similarly, log base two of eight is three\n

work with some fractional values for X. If\n

that saying what power do I raise to two to\n

negative one. It's also easy to compute by\n

base two of 1/4 is negative two since two\n

log base two of one eight is negative f3 I'll\n

pause the video and take a moment to plot\n

one zero, that's here to one that's here,\nfor two

that is here, and then eight, three, which\n

half goes with negative one, and 1/4 with\n

if I connect the dots, I get a graph that\n

and smaller fractions, I would keep getting\n

log base two of them, so my graph is getting\n

more and more negative, as x is getting close\n

graph over here with negative X values, I\n

that omission is no accident. Because if you\n

of a negative number, like say negative four\n

exist because there's no power that you can\n

are no points on the graph for negative X\n

on the graph where x is zero, because you\n

power you can raise to two to get zero. I\n

graph. First of all, the domain is x values\n

can write that as a round bracket because\n

the range is going to be the y values, while\n

of the negative numbers. And the graph gradually\n

So the range is actually all real numbers\n

to infinity. Finally, I want to point out\n

the y axis, that is at the line x equals zero.\n

A vertical asymptote is a line that our functions\n

the graph of y equals log base two of x. But\n

10 of x, it would look very similar, it would\n

zero, a range of all real numbers and a vertical\n

through the point one zero, but it would go\n

log base 10 of 10 is one, it would look pretty\n

But even though it doesn't look like it with\n

goes up to n towards infinity. In fact, the\n

bigger than one looks pretty much the same,\n

we know what the basic log graph looks like,\n

log functions without plotting points. Here\n

five. And again, I'm just going to draw a\n

graph, I probably would want to plot some\n

if it was just like y equals ln of x, that\n

go through the point one zero, with a vertical\n

a graph, ln of x plus five, that just shifts\n

the same vertical asymptote. Since the vertical\n

a vertical line, but instead of going through\n

five. So I'll draw a rough sketch here. Let's\n

x and the transformed version y equals ln\n

and the vertical asymptote. Our original function\n

Since adding five on the outside affects the\n

this transformation doesn't change the domain.\n

Now the range of our original y equals ln\n

Shifting up by five does affect the y values,\n

But since the original range was all real\n

real numbers, you still get the set of all\n

change either. And finally, we already saw\n

y axis x equals zero, when we shift that up\n

x equals zero. In this next example, we're\n

since the plus two is on the inside, that\n

I'll draw our basic log function. Here's our\n

as y equals log of x going through the point\none, zero

here's its vertical asymptote. Now I need\n

asymptote shifts left, and now it's at the\n

x equals zero, and my graph, let's see my\n

negative one zero, since I'm subtracting two\n

the resulting graph. Let's compare the features\n

about domains, the original had a domain of\n

that left. So I've subtracted two from all\n

I can also verify just by looking at the picture.\n

to infinity, well, shifting left only affects\n

range. So my range is still negative infinity\n

at x equals zero. And since I subtract two\n

negative two. In this last problem, I'm not\n

just use algebra to compute its domain. So\n

taking the logs of things? Well, you can't\n

So whatever's inside the argument of the log\n

better be greater than zero. So I'll write\n

than zero. Now it's a matter of solving an\n

3x. So two thirds is greater than x. In other\n

our domain is all the x values from negative\n

thirds. It's a good idea to memorize the basic\n

something like this, go through the point\n

the y axis. Also, if you remember that you\n

zero, then that helps you quickly compute\n

the log function, you set that greater than\n

logs and exponents. Please pause the video\n

evaluate the following four expressions. Remember,\n

log button. While log base e on your calculator\n

that the log base 10 of 10 cubed is three.\n

the log base 10 of 1000 is 1000. And eat the\n

log and the exponential function with the\n

the exponent. In fact, it's true that for\n

equal to x, the same sort of cancellation\n

in the log function with the same base in\n

to the power of log base 10 of 1000, the 10\n

other, and we're left with the 1000s. This\n

a of x is equal to x. We can describe this\n

a log function with the same base undo each\n

of inverse functions, the exponential function\n

these roles hold for the first log role. log\n

power do we raise a two in order to get a\n

is a dx? Well, the answer is clearly x. And\n

notice that the log base a of x means the\n

is saying that we're supposed to raise a to\n

need to raise a two to get x, then we'll certainly\n

examples. If we want to find three to the\n

base three undo each other, so we're left\n

Remember that ln means log base e. So we're\n

functions undo each other, and we're left\n

three z, remember that log without a base\n

we want to take 10 to the log base 10 of three\n

each other. So we're left with a three z.\n

10 to the x equal to x, well, ln means log\n

the x, notice that the base of the log and\n

the same. So they don't undo each other. And\n

usually equal to x, we can check with one\n

e of 10 to the one, that's log base e of 10.\n

equal to 2.3. And some more decimals, which\n

is false, it does not hold. We need the basis\n

each other. In this video, we saw that logs\n

log base a of a to the x is equal to x and\n

for any values of x and any base a. This video\n

log rules are closely related to the exponent\n

the exponent rules. To keep things simple,\n

two. Even though the exponent rules hold for\n

the zero power, we get one, we have a product\n

the M times two to the n is equal to two to\n

two numbers, then we add the exponents. We\n

to the M divided by n to the n is equal to\n

that if we divide two numbers, then we subtract\n

that says if we take a power to a power, then\n

rules can be rewritten as a log rule. The\n

be rewritten in terms of logs as log base\n

base two of one equals zero mean To the zero\n

can be rewritten in terms of logs by saying\n

of y. I'll make these base two to agree with\n

In words, that says the log of the product\n

represent exponent, this is saying that when\n

their exponents, which is just what we said\n

for exponents can be rewritten in terms of\n

equal to the log of x minus the log of y.\n

is equal to the difference of the logs. Since\n

saying the same thing is that when you divide\n

That's how we described the exponent rule\n

can be rewritten in terms of logs by saying\n

log of x. Sometimes people describe this rule\n

with an exponent, you can bring down the exponent\n

power of two, this is really saying when we\n

exponents. That's exactly how we described\n

if you multiply this exponent on the left\n

traditional to multiply it on the left side.\n

but they actually work for any base. To help\n

write out the log roles using a base of a\n

use the log rules to rewrite the following\n

In the first expression, we have a log base\n

of the quotient as the difference of the logs.\n

can rewrite the log of a product as the sum\n

of z. When I put things together, I have to\n

entire log expression. So I need to subtract\n

do that by putting them in parentheses. Now\n

the negative sign. And here's my final answer.\n

product. So I can rewrite that as the sum\nof two logs.

I can also use my power rule to bring down\n

That gives me the final expression log of\n

on this problem is to rewrite this expression\n

those two expressions are not equal. Because\n

whole five times two, we can't just bring\n

all, the power rule only applies to a single\n

to a product like this. And these next examples,\n

we're given sums and differences of logs.\n

log expression. By look at the first two pieces,\n

rewrite it as the log of a quotient. Now I\n

that as the log of a product. I'll clean that\n

five of a times c over B. In my second example,\n

of a product now, I will Like to rewrite this\n

But I can't do it yet, because of that factor\n

the power rule backwards to put that two back\n

So I will copy down the ln of x plus one times\n

as ln of x squared minus one squared. Now\n

logs, which I can rewrite as the log of a\n

more. Since x plus one times x minus one is\n

cancel factors to get ln of one over x squared\n

for logs that are related to exponent rules.\n

one is equal to zero. Second, we saw the product\n

sum of the logs. We saw the quotient rule,\n

the logs. And we saw the power rule. When\n

in it, you can bring down the exponent and\n

no log rule that helps you split up the log\n

is not equal to the sum of the logs. If you\n

together, this kind of makes sense, because\n

of two exponential expressions.

Log rules will be super handy, as we start\n

have an equation like this one that has variables\n

for getting those variables down where you\n

a few examples of solving equations with variables\n

solve for x the equation five times two to\n

I'm going to isolate the difficult spot, the\n

In this example, I can do that by dividing\n

x plus one equals 17 over five. Next, I'm\n

possible to take the log with any base, but\n

base e for the simple reason that my calculator\n

I'll just take the log base 10. So I can omit\n

here. And that gives me this expression. As\n

bring down my exponent and multiply it on\n

here because the entire x plus one needs to\n

third step using the log roles. Now all my\n

I can work with them, but I still need to\n

parentheses. So I'm going to free it from\n

x log two plus log two equals log of 17 fifths.\n

my terms with x's in them to one side, and\n

side. Finally, I factor out my x. Well, it's\n

to isolate it. So I read out what I did. So\n

on one side, and the terms without x's on\n

by factoring out and dividing. We have an\n

maybe not so useful if you want a decimal\n

into your calculator using parentheses liberally\n

always a good idea to check your work by typing\n

checks out. This next example is trickier\n

in two places with two different bases. First\n

by isolating the tricky stuff. But in this\n

or simplifier, or no way to isolate anything\n

to the next step and take the log of both\n

10. But we couldn't use log base e instead.\n

exponents. This gives me 2x minus three in\n

five. Now I'm going to distribute things out\n

gives me 2x log two minus three log two equals\n

to group the x terms on one side, and the\n

side. So I'll keep the 2x log two on the left,\n

and that gives me the minus two log five that\n

on the right. Finally, I need to isolate x\n

mean I factor out the x from all the terms\n

quantity to log two minus log five. And that\n

the right side by the quantity on the left\n

I encourage you to type the whole thing in\n

if you round off, you'll get a less accurate\n

at once. In this example, when I type it in,\n

In this equation, we have the variable t in\n

not a variable, it represents the number e\n

Because there's already an E and the expression,\n

in this problem instead of log base 10. But\n

clean things up. by isolating the tricky parts,\n

And that will give us either the negative\n

T. One way to proceed would now be to clean\n

by either the point two t, but I'm going to\n

take the natural log of both sides. That gives\n

ln of three fifths e to the 0.2 t. Now on\n

rule to bring down my exponent and get minus\n

can't bring down the exponent yet because\n

three fifths. So before I can bring down the\n

using the product rule. So I can rewrite this\n

T. And now I can bring down the exponent.\n

ultimately bring down my exponents. Now ln\n

log base e of E. So that's asking what power\n

answer is one. So anytime I have ln of E,\n

using natural log is a little bit handier\n

that simplification. Next, I'm ready to solve\n

But I do need to bring my T terms to one side\n

let's see. I'll put my T terms on the left\n

And finally I'm going to isolate t by factoring\n

out my T \nand now I can divide. Using my calculator

I can get a decimal answer of 2.04 Three,\n

equations with variables in the exponent.\n

sides and use the log properties to bring\n

examples of equations with logs in them like\n

like this one, we have to free the variable\n

functions. My first step in solving pretty\n

and isolate the tricky part. In this case,\n

it. So I can isolate it by first adding three\n

five equals four, and then I can divide both\n

part, I still need to solve for x, but x is\n

I need to somehow undo the log function. Well\n

each other. And since this is a log base e,\n

e also. So I'm going to take e to the power\n

to the ln 2x plus five, and that's going to\n

Now e to the ln of anything, I'll just write\n

e of a, that you the power and log base e\n

to use that principle over here, e to the\n

e undo each other. And we're left with 2x\n

is equal to E squared. And from there, it's\n

by subtracting five from both sides and then\n

step here is to just say finish solving for\n

solving equations with logs in them. And that's\n

solutions. an extraneous solution is a solution\n

doesn't actually satisfy the original equation.\n

in them, because we might get a solution that\n

zero, and we can't take the log of a negative\n

solution of E squared minus five over two.\n

and see if that works. So let's see the twos\n

five plus five, minus three, I want that to\n

I have two ln e squared minus three that I\n

to work out because let's see, ln is log base\n

the question, What power do I raise e two\n

to the power of two to get e squared. So this\n

that equal one, four minus three does equal\n

have any problem with taking the log of negative\n

solutions. So this is our solution. The second\n

there's a log into places. Now notice that\n

So a base 10 is implied. So I'm already thinking\n

to want to take a 10 to the power of both\n

isolate the tricky part, but there's nothing\n

can't do it here. So we'll just jump right\n

of both sides. Okay, so that's going to give\n

plus log x, that whole thing is in the exponent\n

to do with the right side 10 to the one is\n

while remembering my exponent rules, I know\n

what happens when you multiply two things.\n

x plus three times 10 to the log x, right,\n

add the exponent, so these are the same. Okay,\n

base 10, those undo each other. And so this\n

three. Similarly, 10 to the log base, 10 of\n

by x, that's equal to 10. Now I have an equation\n

going to first multiply out to make it look\n

side. So is equal to zero. And, and now I\n

I think this one factors, it looks like X\n

to get x is negative five, or x is two. So\n

for x. Finally, we need to check our solutions\n

ones. So let's see if x equals negative five.\n

that says, I'm checking that log of negative\n

checking that's equal to one. Well, this is\n

giving you a queasy feeling too, because log\n

can't take the log of a negative number. Same\n

negative five is an extraneous solution, it\n

Let's check the other solution, x equals two.\n

plus three plus log of two is equal to one.\n

of negative numbers, or zero here, this should\n

we can see let's see this is log of five plus\n

my log rules, the sum of two logs is the log\n

we want that to equal one. And that's just\n

one because log base 10 of 10 says, What power\n

is one. So the second solution x equals two\n

Before I leave this problem, I do want to\n

approach. Some people like to start with the\n

to combine everything into one log expression.\n

that's the same as the log of a product, right,\n

x plus three times x, that equals one, then\n

of both sides. And as before, the 10 to the\n

and we get x plus three times x equals 10,\n

we ended up using exponent rules to rewrite\n

we use log rules to rewrite things. So the\n

equivalent, and they certainly will get us\n

examples of equations with logs in them and\n

use exponential functions to undo the log.\n

sides to undo natural log, and take 10 to\n

In this video, we'll use exponential equations\n

growth and radioactive decay. I'll also introduce\n

In this first example, let's suppose we invest\n

interest compounded once a year. How many\n

in it if you don't make any further deposits\n

six point 5% interest each year, that means\n

by 1.065. So after t years, my 1600 gets multiplied\n

function notation as f of t equals 1600 times\n

of money after t years. Now we're trying to\n

$2,000 is a amount of money. So that's an\n

for T the amount of time. So let me write\n

for t. Now to solve for t, I want to first\n

tricky part is the part with the exponential\n

1600. That gives me 2000 over 1600 equals\n

bit further as five fourths. Now that I've\n

going to be to take the log of both sides.\n

And I know that if I log take the log of both\n

down where I can solve for it. I think I'll\n

force equals ln of 1.065 to the T. Now by\n

I can bring that exponent t down and multiply\n

t just by dividing both sides by ln of 1.065.\n

t is approximately 3.54 years. And the next\n

that initially contains 1.5 million bacteria,\n

find the doubling time, the doubling time\n

to double in size. For example, the amount\n

million bacteria to 3 million bacteria would\n

an equation for the amount of bacteria. So\n

in millions, then my equation and T represents\n

by the initial amount of bacteria times the\n

my population of bacteria is growing by 12%\n

gets multiplied by one point 12. Since we're\n

for the t value when P of D will be twice\n

solve for t. As before, I'll start by isolating\n

bringing the T down. And finally solving for\n

write this as ln two over ln 1.12. Using my\n

interesting fact that doubling time only depends\n

initial population. In fact, I could have\n

knowing how many bacteria were in my initial\n

work. If I didn't know how many I started\n

times 1.12 to the t where a is our initial\n

Then if I want to figure out how long it takes\n

with a and double that I get to a. So I'll\n

solve for t. Notice that my A's cancel. And\n

the T down and solve for t, I get the exact\n

the initial population was, I didn't even\n

we're told the initial population, and we're\n

what percent the population increases each\n

multiply the population by each minute. So\n

that I want to use an equation of the form\n

to be the number of minutes, and y is going\n

my initial amount a is 350. So I can really\n

doubling time tells me that when 15 minutes\n

twice as big, or 700. plugging that into my\n

the 15. Now I need to solve for b. Let me\n

sides by 350. That gives me 700 over 350 equals\n

b to the 15th. To solve for B, I don't have\n

in the base not in the exponent, so I don't\n

the easiest way to solve this is just by taking\n

The 1/15 power. That's because if I take B\n

that gives me B to the one is equal to two\n

1/15, which as a decimal is approximately\n

if I'm doing a decimal approximation and these\n

of course, the most accurate thing is just\n

rewrite my equation is y equals 350 times\n

work this problem one more time. And this\n

y equals a times e to the RT. This is called\n

but it's actually an equivalent form to this\n

about why these two forms are equivalent at\n

to solve in this form. So I know that my initial\n

t is 15, my Y is 700. So I plug in 700 here\n

Again, I'm going to simplify things by dividing\n

e to the r times 15. This time my variable\n

want to take the log of both sides, I'm going\n

an E and my problem. So natural log and E\n

then common log with base 10 and E. So I take\n

exponent down. So that's our times 15 times\n

Because Elena V is asking what power do I\n

I get 15 r equals ln two. So r is equal to\n

in to my original equation as e to the ln\n

two equations were actually the same thing\n

that is if I start with this equation, and\n

to the tee. Well, I claim that this quantity\n

the 1/15. And in fact One way to see that\n

Right, that's the same, because every time\n

But what's Ed Oh, and two, E and ln undo each\n

1/15 to the T TA, the equations are really\n

to find two different versions of an exponential\n

b to the T, or the continuous growth one,\n

we're going to work with half life, half life\n

means the amount of time that it takes for\n

we originally started with, we're told that\n

5750 years. So that means it takes that long\n

decay, so that you just have half as much\n

So we're told a sample of bone that originally\n

14 now contains only 40 grams, we're supposed\n

carbon dating. Let's use the continuous growth\n

is our amount of radioactive c 14 is going\n

times e to the RT, we could have used the\n

a times b to the T, but I just want to use\n

that our half life is 5750. So what that means\n

be one half of what we started with. Let me\n

figure out use that to figure out what r is.\n

So I plug in one half a for the final amount,\n

and I have 5750 I can cancel my A's. And now\n

So I do need to take the log of both sides\n

e since I already have an E and my problem,\n

log base 10 is okay. Now, on the left side,\n

of one half on the right side, ln n e to a\n

R times 5750. Now I can solve for r, it's\n

decimal, but it's actually more accurate just\n

my equation, I have f of t equals a times\n

I can use that to figure out my problem. And\n

grams, that's my a, I want to figure out when\n

my final amount. And so I need to solve for\n

by 200. Let's say 40 over 200 is 1/5. Now\n

ln and e to the power undo each other. So\n

divided by five 750 T. And finally I can solve\n

calculator gives me an answer of 13,351 years\napproximately.

That kind of makes sense in terms of the half\n

to decrease by half a little more than two\n

get you to 100 decreasing to half again would\n

half lives is getting pretty close to 13,000\n

things it introduced continue Less growth\n

writing an exponential function. The relationship\n

thing as EDR. In that version, it also introduced\n

amount of time it takes a quantity to double\n

model. Recall that a linear equation is equation\n

an equation without any x squared or y squared\n

the form y equals mx plus b, the equation\n

a collection of two or more linear equations.\n

A solution to a system of equations is that\n

of the equations. For example, the ordered\n

equals three is a solution to this system.\n

three into the first equation, it checks out\n

one. And if I plug in x equals two and y equals\n

out two plus three equals five. However, the\n

y equals four is not a solution to the system.\n

second equation, since one plus four does\n

because two times one minus four is not equal\n

methods to find the solutions to systems of\n

want to solve this system of equations, there\n

use the method of substitution, or we could\n

method of substitution, the main idea is to\n

then substitute it in to the other equation.\n

3x minus two y equals four, and isolate the\n

dividing both sides by three. Think I'll rewrite\n

into two fractions four thirds plus two thirds\n

equation 5x plus six y equals two. And I'm\n

That gives me five times four thirds plus\n

I've got an equation with only one variable\n

First, I'm going to distribute the five so\n

six y equals two. And now I'm going to keep\n

on the left side, but I'll move all my terms\n

this point, I could just add up all my fractions\n

like working with fractions, I think I'll\n

here. So I'm going to actually multiply both\n

to get rid of the denominators and not have\n

down. Distributing the three, I get 10 y plus\n

things together. So that's 28 y equals negative\n

14 over 28, which is negative one half.

So I've solved for y. And now I can go back\n

solve for x, I plug it into my first equation.\n

That gives me 3x plus one equals four. So\n

to one. I've solved my system of equations\n

one half, I can also write that as an ordered\n

Now let's go back and solve the same system,\n

elimination, the key idea to the method of\n

a constant to make the coefficients of one\n

my two equations. Say I'm trying to make the\n

is to multiply the first equation by five,\n

the coefficient of x will be 15 for both equations,\n

I'm going to multiply both sides by five.\n

multiply both sides by three. That gives me\n

20. And for the second equation, 15x plus\n

the coefficients of x match. So if I subtract\n

term will completely go away, it'll be zero\n

10 y minus 18, y is going to give me minus\n

to give me 14. solving for y, I get y is 14\n

like before. Now we can continue, like we\n

that value of y into either one of the equations.\n

proceeds as before. So once again, I get the\n

one half. Before we go on to the next problem,\n

Here I've graphed the equations 3x minus two,\n

And we can see that these two lines intersect\n

one half, just like we predicted by solving\n

at another system of equations. I'm going\n

is on the left side with the y term, and the\n

rewrite or copy down the second equation.\n

is minus four, and the second equation is\n

elimination and multiply the first equation\n

That'll give me a coefficient of x of negative\n

second equation, those are equal and opposite,\n

equations to cancel out access. So let's do\n

plus 24, y equals three, and I'll put everything\n

everything by four. So that's 12x minus 24,\n

has happened here, not only do the x coefficients\n

Y coefficients do also. So if I add together\n

x term, I'm also going to cancel out the y\n

equal to three plus eight is 11. Well, that's\n

to 11. And that shows that these two equations\n

Let's look at this situation graphically.\n

they're parallel lines with the same slope.\n

each equation, the first equation, I get y\n

the same thing as four eighths or one half\n

if I isolate y, let's say minus six y equals\n

y equals one half x minus 1/3. So indeed,\n

with different In intercepts, and so they\n

sense that we have no solution to our system\n

has no solution is called an inconsistent\n

behavior happens. This time, I think I'm going\n

have X with a coefficient of one. So it's\n

equation, and then plug in to the second equation\n

y equals 18. If I distribute out, I get the\n

and I just get 18 equals 18, which is always\n

of linear equations. If you look more closely,\n

just a constant multiple, the first equation\n

as big as the corresponding term and the first\n

the second equation, anything, any x and y\n

the second one. So this system of equations\n

pair x y, where X plus five y equals six,\n

will satisfy this system of equations. That\n

x value of six or a y value of one corresponding\n

Corresponding to an x value of 13 thirds just\n

Graphically, if I graph both of these equations,\n

so I'll just see one line. In this video,\n

using the method of substitution and the method\n

linear equations can have one solution. When\n

in one point, they can be inconsistent, and\n

lines, or they can be dependent and have infinitely\n

lying on top of each other. In this video,\n

rate and time. The key relationship to keep\n

distance traveled divided by the time it takes\n

60 miles an hour, that's your rate. And that's\n

in one hour. Sometimes it's handy to rewrite\n

by T time. And that gives us that R times\n

is equal to rate times time. There's one more\n

the idea that rates add. For example, if you\n

you're walking on a moving sidewalk, that's\n

your total speed of travel with respect to\n

be three plus two, or five miles per hour.\n

the first rate per second rate is equal to\n

distance equals rate times time, and rates\n

has a top speed of six miles per hour and\n

top speed. She went 10 miles upstream in the\n

we're supposed to find the rate of the river\n

in this problem into a chart.

During the course of Elsa stay, there were\n

one period of time she was going upstream.\n

downstream. For each of those, I'm going to\n

she went at and the time it took when she\n

of 10 miles. When she was going downstream\n

the times to travel those two distances were\n

was, I'll just give it a variable I'll call\n

rate she traveled upstream was slower because\n

when she was going downstream with the current.\n

is, that's what we're trying to figure out.\n

do know that in still water also can go six\n

since she's going with the direction of the\n

should be six plus R, that's her rate, and\n

On the other hand, when she's going upstream,\n

rate of six miles per hour, we need to subtract\n

we've charted out our information, we can\n

distance equals rate times time, we actually\n

T, and 30 is equal to six plus R times T.\n

a system of equations, our next job is to\n

I think the easiest way to proceed is to isolate\n

first equation, I'll divide both sides by\n

by six plus R. That gives me 10 over six minus\n

t. Now if I set my T variables equal to each\n

equal to 30 over six plus R. I'm making progress\n

single variable that I need to solve. Since\n

I'm going to proceed by clearing the denominator.\n

least common denominator, that is six minus\n

I get that the six plus r times 10 is equal\n

I'm going to get 60 plus xR equals 180 minus\n

going to be 40 r is equal to 120. So our,\n

miles per hour. This is all that the problem\n

wanted to solve for the other unknown time,\n

my equations and solving for T. In this video,\n

charting out my information for the two situations\n

to fill in some of my boxes, and then using\n

to build a system of equations. In this video,\n

we have to figure out what quantity of two\n

contains 6% sodium hypochlorite. The other\n

be combined with 70 liters of a weaker 1%\n

that's 2.5% sodium hypochlorite. I want to\n

So I'm asking myself what quantities are going\n

amount of sodium hypochlorite that has symbol\n

it should equal the total amount of sodium\n

amount of water before mixing should equal\n

there's just the total amount of solution.\n

with water should equal the total amount of\n

I'm looking for. But before I start reading\n

out my quantities. So I've got the 6% solution.\n

And I've got my Desired Ending 2.5% solution.\n

certain volume of sodium hypochlorite. I've\n

total volume of solution. Let me see which\n

that I'm adding 70 liters of the 1% solution.\n

here. I don't know what volume of the household\n

to find out. So I'm going to just call that\n

by combining my other two solutions, I know\n

two volumes, so I'll write 70 plus x in this\n

the volume of solution is, 6% of that is the\n

sodium hypochlorite is going to be 0.06 times\n

whatever's left, so that's going to be x minus\n

following the same reasoning for the 1% solution\n

So that's going to be 0.01 times 70. Or point\n

is going to be 99% or point nine, nine times\n

for the 2.5% solution, the volume of the sodium\n

the volume of solution 70 plus x and the volume\n

that's 0.975 times 70 plus x. Now I've already\n

before added up is the volume of solution\n

But I haven't yet used the fact that the volume\n

and after. So I can write that down as an\n

equal to 0.025 times 70 plus x. Now I've got\n

it. Since I don't like all these decimals,\n

by let's see, 1000 should get rid of all the\n

700 equals 25 times 70 plus x. Distributing\n

1750 plus 25x. So let's see 60 minus 25 is\n

x equals 30 liters of the household bleach.

Notice that I never actually had to use the\n

mixing is equal to the quantity of water after\n

column. In fact, that information is redundant.\n

hypochlorite add up, and the total volume\n

of waters add up is just redundant information.\n

involving solutions can be used to solve many\n

items. My favorite method is to first make\n

the types of items in your mixture. Fill in\n

fact that the quantities add. This video is\n

Recall that a rational function is a function\n

of two power. No Here's an example. The simpler\n

considered a rational function, you can think\n

graph of this rational function is shown here.\n

of a polynomial. For one thing, its end behavior\n

is the way the graph looks when x goes through\n

numbers, we've seen that the end behavior\n

cases. That is why marches off to infinity\n

big or really negative. But this rational\n

Notice, as x gets really big, the y values\nare leveling off

at about a y value of three. And similarly,\n

is leveling off near the line y equals three,\n

graph, that line is called a horizontal asymptote.\n

that our graph gets closer and closer to as\n

infinity, or both. There's something else\n

graph, look at what happens as x gets close\n

five with x values on the right, our Y values\n

And as we approach the x value of negative\n

up towards positive infinity. We say that\n

negative five. A vertical asymptote is a vertical\n

to. Finally, there's something really weird\n

open circle there, like the value at x equals\n

is a place along the curve of the graph where\n

identified some of the features of our rational\nfunctions graph

I want to look back at the equation and see\n

just by looking at the equation. To find horizontal\n

is doing when x goes through really big positive\n

our equation for our function, the numerator\n

term when x is really big, right, because\n

enormous compared to this negative 12. If\n

the denominator, the denominator will be dominated\n

big positive or negative number, like a million,\n

than three times a million or negative 10.\n

or the horizontal asymptote, for our function,\n

and the term on the denominator that have\n

dominate the expression in size. So as x gets\n

to be approximately 3x squared over x squared,\n

asymptote at y equals three. Now our vertical\n

denominator of our function is zero. That's\n

denominator is zero. And when we get close\n

we're going to be dividing by tiny, tiny numbers,\n

magnitude. So to check where our denominators\n

I'm going to go ahead and factor the numerator\n

let's see, pull out the three, I get x squared\n

factors into X plus five times x minus two,\n

further, that's three times x minus two times\n

when x is equal to negative five, my denominator\n

zero. That's what gives me the vertical asymptote\n

x equals two, the denominator is zero, but\n

cancelled the x minus two factor from the\n

form for my function that agrees with my original\n

That's because when x equals two, the simplified\n

does not, it's zero over zero, it's undefined.\n

near x equals to our original functions just\n

our function only has a vertical asymptote\n

two, because the x minus two factor is no\n

it does have a hole at x equals two, because\n

even though the simplified version is if we\n

just plug in x equals two into our simplified\n

of three times two plus two over two plus\n

thirds. So our whole is that to four thirds.\n

detail, let's summarize our findings. We find\n

where the denominator is zero. The holes happen\n

zero and those factors cancel out. The vertical\n

denominator is zero, we find the horizontal\n

term on the numerator and the denominator,\n

three examples. In the first example, if we\n

to 5x over 3x squared, which is five over\n

is going to be huge. So I'm going to be dividing\n

to be going very close to zero. And therefore\n

at y equals zero. In the second example, the\n

simplifies to two thirds. So as x gets really\n

thirds, and we have a horizontal asymptote\n

the highest power terms, x squared over 2x\n

big, x over two is getting really big. And\n

at all. This is going to infinity, when x\n

and is going to negative infinity when x goes\n

case, the end behavior is kind of like that\n

asymptote. In general, when the degree of\n

the denominator, we're in this first case\n

to the numerator and we go to zero. In the\n

and the degree of the dominant are equal,\n

asymptote at the y value, that's equal to\n

in the third case, when the degree of the\n

denominator, then the numerator is getting\n

we end up with no horizontal asymptote. Final\n

to one more example. Please pause the video\n

horizontal asymptotes and holes for this rational\n

and holes, we need to look at where the denominator\n

factor both the numerator and the denominator.\n

we might have a whole instead of a vertical\n

factor. Let's see that's 3x times x plus one\n

x. And then I'll factor some more using a\n

a 2x and an X to multiply together to the\n

one or alpha minus three and a one. Let's\n

one times x plus three that does get me back\n

checks out. Now I noticed that I have a common\n

denominator. So that's telling me I'm going\n

I could rewrite my rational function by cancelling\n

as long as x is not equal to zero. So the\n

zero into my simplified version, that would\n

zero minus one times zero plus three, which\n

So my whole is at zero minus one. Now all\n

make my denominator zero will get me vertical\n

when 2x minus one times x plus three equals\n

or x plus three is zero. In other words, when\n

Finally, to find my horizontal asymptotes,\n

term in the numerator and the denominator.\n

bottom heavy, right? When x gets really big,\n

means that we have a horizontal asymptote\n

of our graph, the whole, the vertical asymptotes\n

would give us a framework for what the graph\n

at y equals zero, vertical asymptotes at x\n

at a hole at the point zero minus one. plotting\n

of graphing program, we can see that our actual\n

Notice that the x intercept when x is negative\n

our rational function or reduced rational\n

a zero on the numerator that doesn't make\n

zero. And an X intercept is where the y value\n

we learned how to find horizontal asymptotes\n

highest power terms, we learned to find the\n

at the factored version of the functions.\n

make the numerator and denominator zero, his\n

asymptotes correspond to the x values that\n

any any common and in common factors in the\n

This video is about combining functions by\n

dividing them. Suppose we have two functions,\n

x squared. One way to combine them is by adding\n

x means the function defined by taking f of\n

that means we take x plus one and add x squared,\n

plus x plus one. So f plus g evaluated on\n

I wanted to evaluate f plus g, on the number\n

one, or seven. Similarly, the notation f minus\n

f of x and subtracting g of x. So that would\n

to take f minus g evaluated at one, that would\n

the notation F dot g of x, which is sometimes\n

we take f of x times g of x. In other words,\n

simplified as x cubed plus x squared. The\n

f of x and divided by g of x. So that would\n

figure, the blue graph represents h of x.\n

p of x, we're asked to find h minus p of zero.

We don't have any equations to work with,\n

minus p of x is defined as h of x minus p\n

is going to be h of zero minus p of zero.\n

finding the value of zero on the x axis, and\n

function h of x. So that's about 1.8. Now\n

for zero on the x axis, and finding the corresponding\n

a y value of one 1.8 minus one is 0.8. So\n

of zero. If we want to find P times h of negative\n

negative three times h of negative three.\n

value of negative three, the y value for P\n

corresponds to a y value of negative two for\nH.

Two times negative two is negative four. So\n

In this video, we saw how to add two functions,\n

and divide two functions in the following\n

the first function, and then you apply the\n

function. For example, the first function\n

years. So its input would be time in years,\n

of people in the population. The second function\n

of population size. So it will take population\n

costs. If you put these functions together,\n

way from time in years to healthcare costs.\n

F. The composition of two functions, written\n

as follows. g composed with f of x is G evaluated\n

and diagram f x on a number x and produces\n

f of x and produces a new number, g of f of\n

with F is the function that goes all the way\n

examples where our functions are defined by\n

with F of four, by definition, this means\n

we always work from the inside out. So we\n

f of four, using the table of values for f\n

so we can replace F of four with the number\n

seven becomes our new x value in our table\n

to the G of X value of 10. So g of seven is\n

F of four is equal to 10. If instead we want\n

can rewrite that as f of g of four Again work\n

g of four. So four is our x value. And we\n

g of four is one. So we replaced you a four\n

Using our table for F values, f of one is\n

f of four, we got a different answer than\n

g composed with F is not the same thing as\n

and take a moment to compute the next two\n

of two by the equivalent expression, f of\n

know that f of two is three, and f of three\n

g of six, rewrite that as f of g of six, using\n

of eight, eight is not on the table as an\n

there is no F of eight, this does not exist,\n

for F composed with g. Even though it was\n

the way through and get a value for F composed\n

to the composition of functions that are given\nby equations.

p of x is x squared plus x and q of x as negative\n

As usual, I can rewrite this as Q of P of\n

is one squared plus one, so that's two. So\n

of two is negative two times two or negative\n

In this next example, we want to find q composed\n

as usual as Q of p of x and work from the\n

for that. That's x squared plus x. So I can\n

I'm stuck with evaluating q on x squared plus\n

that thing. So q of x squared plus x is going\n

plus x, what I've done is I've substituted\n

where I saw the X in this formula for q of\n

So that will be multiplying negative two by\n

piece, I can simplify this a bit as negative\n

for Q composed with p of x. Notice that if\n

which I already did in the first problem,\n

two times one squared minus two and I get\n

try another one. Let's try p composed with\n

x. Working from the inside out, I can replace\n

P of negative 2x. Here's my formula for P.\n

plug in this expression everywhere I see an\n

2x squared plus negative 2x. Again, being\n

plug in the entire expression in forex. let\n

Notice that I got different expressions for\n

we see that q composed with P is not necessarily\n

video and try this last example yourself.\n

we're going to replace p of x with its expression\n

p on x squared plus x. That means we plug\n

x in this formula, so that's x squared plus\n

Once again, I can simplify by distributing\n

cubed plus x squared plus x squared plus x,\n

plus x. In this last set of examples, we're\n

for a function of h of x. But we're supposed\n

functions, F and G. Let's think for a minute,\n

first, f composed with g of x, let's see,\n

these expressions from the inside out, we\n

to figure out what what f and g could be,\n

my expression for H, so I'm going to draw\n

inside the box, that'll be my function, g\n

whatever happens to the box, in this case,\n

my outside function, my second function f.

So here, we're gonna say g of x is equal to\n

to the square root of x, let's just check\n

check that when I take the composition, f\n

as my original h. So let's see, if I do f\n

that's f of g of x, working from the inside\n

x squared plus seven. So I need to evaluate\n

in x squared plus seven, into the formula\n

of x squared plus seven to the it works because\n

a correct answer a correct way of breaking\n

But I do want to point out, this is not the\n

for H of X again, and this time, I'll put\n

the x squared. If I did that, then my inside\n

be x squared. And my second function is what\n

to the box, and the box gets added seven to\n

words, f of x is going to be the square root\n

works. If I do f composed with g of x, that's\n

I'm taking f of x squared. When I plug in\n

root of x squared plus seven. So this is that\n

we learn to evaluate the composition of functions.\n

out. We also learn to break apart a complicated\n

by boxing one piece of the function and letting\n

Let that be the inside of the box, and the\n

be whatever happens to the box.

The inverse of a function undoes what the\n

shoes would be to untie them. And the inverse\n

would be the function that subtracts two from\n

their properties. Suppose f of x is a function\n

two is three, f of three is five, f of four\n

function for F written f superscript. Negative\n

three, F inverse takes three, back to two.\n

of three is to. Similarly, since f takes three\n

since f takes four to six, f inverse of six\n

inverse of one is five. I'll use these numbers\n

of values when y equals f of x and the chart\n

closely related. They share the same numbers,\n

the y values for f inverse of x, and the y\n

for f inverse of x. That leads us to the first\n

of y and x. I'm going to plot the points for\n

points for y equals f inverse of x in red.\n

kind of symmetry you observe in this graph.\n

points, you might have noticed that the blue\n

over the mirror line, y equals x. So our second\n

of x can be obtained from the graph of y equals\nf

by reflecting over the line y equals x. This\n

roles of war annex. In the same example, let's\n

means composition. In other words, we're computing\n

the inside out. So that's f inverse of three.\n

three, we see as to similarly, we can compute\n

take f of f inverse of three. Since f inverse\n

computing F of two, which is three. Please\n

other compositions. You should have found\n

of f of a number, you get back to the very\n

if you take f of f inverse of any number,\n

with. So in general, f inverse of f of x is\n

equal to x. This is the mathematical way of\n

Let's look at a different example. Suppose\n

a moment, and guess what the inverse of f\n

work that F does. You might have guessed that\n

function, we can check that this is true by\n

the cube root of function, which means the\ncube root function

cubed, which gets us back to x. Similarly,\n

And we get back to excellence again. So the\n

the cubing function. When we compose the two\n

we started with. It'd be nice to have a more\n

besides guessing and checking. One method\n

of y and x. So if we want to find the inverse\n

x over 3x. We can write it as y equals five\n

x To get x equals five minus y over three\n

multiply both sides by three y. Bring all\n

alternate without wizened them to the right\n

why this gives us f inverse of x as five over\n

f and our inverse function, f inverse are\n

reciprocals of each other. And in general,\n

over f of x. This can be confusing, because\n

mean one of our two, but f to the minus one\n

reciprocal. It's natural to ask us all functions\n

you might encounter. Is there always a function\n

is no. See, if you can come up with an example\n

function. The word function here is key. Remember\n

x values and y values, such that for each\n

y value. One example of a function that does\n

the inverse of this function is not a function.\n

number two and the number negative two, both\n

you would have to send four to both two and\n

it might be easier to understand the problem,\n

Recall that inverse functions reverse the\n

line y equals x. But when I flipped the green\n

red graph. This red graph is not the graph\n

line test. The reason that violates the vertical\n

violates the horizontal line test, and has\n

a function f has an inverse function if and\n

line test, ie every horizontal line intersects\n

video for a moment and see which of these\n

In other words, which of the four corresponding\n

You may have found that graphs A and B violate\n

would not have inverse functions. But graph\n

So these graphs represent functions that do\n

horizontal line test are sometimes called\n

is one to one, if for any two different x\n

of x one and f of x two are different numbers.\n

whenever f of x one is equal to f of x two,\n

example, let's try to find P inverse of x,\n

two drawn here. If we graph P inverse on the\n

graph simply by flipping over the line y equals\n

we can write y equal to a squared of x minus\n

for y by squaring both sides adding two. Now\n

two, that would look like a parabola, it would\n

together with another arm on the left side.\n

consists only of this right arm, we can specify\n

that x has to be bigger than or equal to zero.\n

graph for the square root of x, y was only\n

closely at the domain and range of P and P\n

values of x such that x minus two is greater\n

the square root of a negative number. This\n

or equal to two, or an interval notation,\n

of P, we can see from the graph is all y value\n

from zero to infinity. Similarly, based on\n

is x values greater than or equal to zero,\n

range of P inverse is Y values greater than\n

to infinity. If you look closely at these\n

domain of P corresponds exactly to the range\n

This makes sense, because inverse functions\n

f inverse of x is the x values for F inverse,\n

of F. The range of f inverse is the y values\n

or the domain of f. In this video, we discussed\n

inverse functions, reverse the roles of y\n

x is the graph of y equals f of x reflected\n

F with F inverse, we get the identity function\n

f inverse with F, that brings x to x. In other\n

function f of x has an inverse function if\n

the horizontal line test. And finally, the\n

the range of f is the domain of f inverse.\n

be important when we study exponential functions\n