Find all vertical asymptotes of f(x)= x / x^2 -9?

0 ⤊

2007-06-13 06:32:28 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

Set the denominator to 0:

0 = x^2 - 9
x^2 = 9
x = +/- 3

So, the asymptotes are -3 and 3.

2007-06-13 06:37:02 · answer #1 · answered by yeeeehaw 5 · 0⤊ 0⤋

In this problem, a vertical asymptote occurs anywhere where the denominator is 0.

x^2 - 9 = 0
x^2 = 9

x = +-3

2007-06-13 13:37:09 · answer #2 · answered by ryanker1 4 · 0⤊ 0⤋

f(x) = x / ( (x - 3).(x + 3) )
Vertical asymptotes are x = 3 , x = - 3

2007-06-13 13:41:13 · answer #3 · answered by Como 7 · 0⤊ 0⤋

Do your own homework. Damn internet.

2007-06-13 13:46:20 · answer #4 · answered by kivesta 1 · 0⤊ 0⤋