How do you graph $(2x^2) /( x^2 - 9)$ ?

Question

3952 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-16T20:52:46

$y = (2x^2)/(x^2-9) = (2(x^2-9+9))/(x^2-9) = 2 + 18/((x+3)(x-3)) = 2 - 3/(x+3) + 3/(x-3)$

Vertical asymptotes:
$x = 3$ and $x = -3$
$lim_(x rarr 3^(+-)) = +-oo$
$lim_(x rarr -3^(+-)) = ""_+^(-)oo$

Horizontal asymptotes:
$y = 2$
$lim_(x rarr +- oo) = 2^+$

Stationary points:
$y' = 3/(x+3)^2 - 3/(x-3)^2 = 0$
$x = 0, y = 0$

$y'' = -6/(x+3)^3 + 6/(x-3)^3$
$y''|_(x=0) = -4/9 < 0,$ therefore local maxima

graph{(2x^2)/((x-3)(x+3)) [-6, 6, -10 10]}

How do you graph (2x^2) /( x^2 - 9)(2x^2) /( x^2 - 9)?