# The function f(x) = x / (x2 - 4) has asymptotes at?

The function f(x) = x / (x2 - 4) has asymptotes at?

Relevance

at x = +/- 2

f(x) = x / (x^2 - 4)

= x / (x-2)(x+2)

because the denominator is the difference of two squares

Asymptotes occur particularly when the denominator = 0 and therefore x = 2 or -2

Alternatively

x^2 - 4 = 0 using your original denominator

x^2 = 4

x = +/- 2

and y does not = 0 at the asymptote ... oh - I see what they mean. Horizontal asymptote at 0 as x tends towards infinity

Asymptotes are parts of the graph where the function is undefined because of division by zero, so set the denominator = 0 and solve

x^2 - 4 = 0

x^2 = 4

x = +/- 2

if I assume that it is x^2 then the answer is +2 and - 2. these two numbers will make 0 in the denominator which would be the asymptote.

If you meant 2x, then the answer is just 2.

• TFV
Lv 5

1) x = 2 or -2 [makes the denominator zero]

2) y = 0 [f(x) approaches 0 as x approaches + or - infinity]