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All right, what I'd like to do is

show you guys how to find the vertical and horizontal

asymptotes.

Remember what an asymptote is it's

as our graph is approaching a certain number

it's going to keep on approaching that line,

but it's never actually going to go ahead and get to that point.

So on a graph as our values keep on getting larger,

it's going to keep on getting closer to the line.

But that's what we also like to call it just [INAUDIBLE]

approaches infinity.

So what I want to do here is, if I

want to find the vertical asymptote

for both of these equations what I need to do

is I need to set both of my denominators to 0.

So I'd say 0 is equal to x minus 2 cubed.

OK, and then here I get 0 equals x squared plus 1.

So when I go ahead and do this, I go and factor this,

I get x equals 2.

OK, I take a cubed root on both sides.

Do you guys need me to show that?

We OK with that?

Take a cube root.

I'll just show it.

Cube root on both sides.

To get rid of the cubed, therefore, then you

have 0 is equal to x minus 2.

Add 2 on both sides.

x is equal to 2, OK?

Over here I notice that there is no number.

That's going to be an imaginary number, right?

That's going to be i.

Square root over 1 equals x, right?

We worked on taking the i.

Because that's going to be negative.

You take root it's going to be the square root of negative 1.

So therefore, there is no x real There's

no x-intercepts or vertical asymptotes for here.

So that's your vertical.

And here vertical we have none.

I don't know why I did a double e.

And now we go ahead and find the horizontal asymptote.

Remember the horizontal asymptote, what we need to do

is we need to look at your exponents

of your leading coefficient.

And if we have our degrees up here is what we look at.

And here you can pretty much write we have x to the 0,

right?

Because there's nothing written there.

So it's x to the 0.

So if you guys remember what we wrote down

last class period is, when you go ahead and compare this

eventually if I FOILed this all out I'm

going to get x cubed, right?

So essentially, what I have is my two numbers, x over x cubed.

And if you guys remember, that whenever

the degree in the numerator was less

than the degree in the denominator, therefore,

you had a horizontal asymptote of 0.

So therefore, horizontal equals y equals 0.

All right, and then for number 12 now

we notice that we have x squared is equal to x squared, right?

So we have when I'm looking at my leading coefficients,

now these are equal to each other.

So when they're equal to each other, we have a

and b is what we call their coefficients of them.

And what they do is when you have

2 is equal to 2, when you have their degrees

are equal to each other, then the horizontal asymptote

is the ratio of a over b.

So the horizontal for this problem,

is going to be 3 over 1.

Because the coefficient for this problem is 1.

So it would be 3 over 1.

So that is how you find the vertical and horizontal

asymptotes of given functions.