How do you find the asymptotes for f(x) = (2x^2 - 8) / (x^2 - 16)?

Feb 4, 2016

vertical asymptotes at x = ± 4
horizontal asymptote at y = 2

Explanation:

Vertical asymptotes occur as the denominator of a rational function approaches zero. To find the equation let the denominator equal zero.

solve ${x}^{2} - 16 = 0$

This is a difference of squares and factors as (x-4)*x+4)=0

the equation of the vertical asymptotes are x = ± 4

[horizontal asymptotes occur as  lim_(x→±∞) f(x) → 0]

If the degree of the denominator and numerator are equal then the equation can be found by taking the ratio of leading coefficients.

In this question they are both of degree 2

hence equation is $y = \frac{2}{1} = 2$

Here is the graph of f(x) to illustrate these.
graph{(2x^2-8)/(x^2-16) [-10, 10, -5, 5]}