Question

Find the vertical asymptotes, if any, and the horizontal or oblique asymptote, if any, of y= (x^2 -9)/(x^2 -4)

The vertical asymptote are x=_____
OR
There is no vertical asymptotes

The horizontal asymptotes are y=______
OR
There are no horizontal asymptotes

Answer 1

vertical asymptotes are values of x where the denominator = zero

there are 2 vertical asymptotes

Answer 2

as function ---> 1 x -->> +/- infinity so what is horizontal asymptote?

Answer 3

this should help

Answer 4

Confuse...

Answer 5

can you factor the denominator..?

Answer 6

vertical asymptote:
x^2 - 4 = 0 - solve this bay factorising to give 2 values of x

Answer 7

okay so it will be (x-3)(x+3)/(x-2)(x+2)=y

Answer 8

the simply way to find the oblique or horizontal asymptote ( in this case) is to simplify

\[\frac{x^2}{x^2} = ?\]

normally it takes a little polynomial division.... but can be found by looking at the leading terms...

Answer 9

thats correct..

so the denominator can't be zero so the vertical asymptotes occur when

x - 2 = 0

and x + 2 = 0


solve for x to find the asymptotes

Answer 10

it will be 2 and -2

Answer 11

thats correct... now you have the vertical asymptotes.

go bale and simplify for the horizontal...

Answer 12

wait how to find for horizontal?

Answer 13

plug in for x right?

Answer 14

you can use polynomial division or in this question because the leading terms have the same power just simplify

\[\frac{x^2}{x^2} = ?\]

whats the value of ?

Answer 15

I got 11/6 and 7/2 when t plug in x=-2 and 2 into the original equation.

Answer 16

would that be right?

Answer 17

when numerator and denominator's leading term have the same degree, the horizontal asymptote would be at their coefficient's division -> n/d

Answer 18

don't plug in any values.... you don't need to just whats the simplified value of

\[\frac{x^2}{x^2} = 1\]

so the oblique or horizontal asymptote is x = 1

Answer 19

okay so do you use that equation to find for the horizontal. would it always = to 1

Answer 20

nope... it requires the numerator to be divided by the denominator... to find the oblique asymptote...

its just in this question its nice and neat and equal to 1.

Answer 21

there should be an oblique asymptote, if the numerator degree were 1 higher than the denominator :S, in this case, they're both the same

Answer 22

confuse so its not 1 for the oblique?

Answer 23

no, it means it has no oblique asymptote

Answer 24

here is what you need

vertical asymptotes at x = -2, 2

horizontal asymptote at y = 1

thats all you need

Answer 25

so as campbell_st said, 1 is correct, that is y=1, but that's just a straight line, no an oblique one

Answer 26

its oblique with a slope of zero....

Answer 27

okay thank you guys for clearing it for me!