Question

Give an example of a rational function that has a horizontal asymptote at y = 0 and a vertical asymptote at,
x = 2 and x = 1.

Answer 1

Alright, when does an equation have a vertical asymptote?

Answer 2

when the bottom equals zero

Answer 3

correct. So if we want there to be a vertical asymptote at x=2 and x=1, we need an equation that has the bottom equal to zero when x=2 and x=1. Can you think of any?

Answer 4

no

Answer 5

How about (x-2)(x-1)?

Answer 6

oh is that the answer?

Answer 7

That's the bottom part of the rational function, because if you plug in either x=2 or x=1, you get zero. Understand?

Answer 8

yeah

Answer 9

So now we have \[\frac{ a }{(x-2)(x-1) }\]

Answer 10

When is there a horizontal asymptote?

Answer 11

idk

Answer 12

at zero

Answer 13

For this problem, correct. There is a horizontal asymptote at y=0. Unlike vertical asymptotes, horizontal asymptotes can be crossed. A horizontal asymptote mainly indicates long term behavior a.k.a the limit of a function

Answer 14

If the hor. asymptote= 0, then that means as x-> infinity and x-> -infinity, y-> 0

Answer 15

ok so the answer is 0/x-2)(x-1

Answer 16

No, not quite. If you have just zero in the numerator, then y always equals 0.

Answer 17

You want to make the function have long term behavior of what I described above.

Answer 18

so like 1-1

Answer 19

To do that, you take the leading coefficients (the first numbers of the top and the bottom) and divide them by each other.

Answer 20

so... right now we have \[\frac{ a }{ x^2}\]

Answer 21

yeah

Answer 22

because x^2 is the first number if we expand (x-1)(x-2)

Answer 23

what is a?

Answer 24

a is the top part of the function that we are still trying to find.

Answer 25

yeah what would it be how do you find it

Answer 26

\[\frac{ a }{ x^2}=0\]

Answer 27

x+1)(x+2?

Answer 28

a=x+1)(x+2?

Answer 29

That's an idea, but if we had that in the top then the leading coefficient becomes x^2

Answer 30

Then we would have \[\frac{ x^2 }{ x^2 } \]

Answer 31

ok so then what is it i dont get it

Answer 32

It's pretty simple I'm just having a hard time explaining it. The top can just be x

Answer 33

oh ok

Answer 34

If we just have x, then we get \[\frac{ x }{ x^2}\]

Answer 35

which as x gets really big equals 0

Answer 36

so final answer is \[\frac{ x }{ (x-2)(x-1)}\]