Question

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

Answer 1

`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]}