How to find the asymptotes of $f(x) =(9x^2 – 36)/( x^2 - 9)$ ?

Question

2260 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-11T00:38:20

Vertical asymptotes occur when the denominator equals $0$ ...

Vertical Asymptotes :
$x^2-9=0$
$x^2=9$
$x=3 " or " x=-3$

Horizontal Asymptotes :
Find the highest exponents in the numerator and denominator, then divide ...

$y=(9x^2)/x^2=9$

So, $y=9$ is the horizontal asymptote.

Here is a graph. The dotted lines are the asymptotes ...

Hope that helped

enter image source here

How to find the asymptotes of f(x) =(9x^2 – 36)/( x^2 - 9)f(x) =(9x^2 – 36)/( x^2 - 9) ?