How do you find the vertical, horizontal or slant asymptotes for #(2x^2 - 8) / (x^2 - 16)#?

1 Answer
Mar 22, 2016

Answer:

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

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s equate the denominator to zero.

solve: # x^2 - 16 = 0 → (x-4)(x+4) = 0 → x = ± 4 #

Horizontal asymptotes occur as #lim_(x→±∞) f(x) → 0 #
divide all terms on numerator and denominator by # x^2 #

#((2x^2)/x^2 - 8/x^2)/(x^2/x^2 - 16/x^2) = (2 - 8/x^2)/(1 - 16/x^2)#

As x →∞ # 8/x^2" and " 16/x^2 → 0 #

# rArr y = 2/1 = 2 " is the asymptote " #

Slant asymptotes occur when the degree of the numerator is greater than the degree of the denominator , which is not the case here , so there are no slant asymptotes.

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