How do you find the asymptotes for #g(x) = (4x^2+11) /( x^2+8x−9)#?

Question

Chloe Alexander · Accepted Answer

H.A. @ #x=0#
V.A.s @ #x=-9 and x=1#
No S.A.

Original Equation: #g(x)=(4x^2+11)/(x^2+8x-9)# Factor: #g(x)=(4x^2+11)/((x+9)(x-1)# The rules for horizontal asymptotes: Because the degrees in this equation are the same, there is a horizontal asymptote at #x=0# The rule for vertical asymptotes: Because #-9 and 1# will cause the equation to be undefined (the denominator is equal to #0#), there are vertical asymptotes at #x=-9 and x=1# The rule for slant asymptotes: Because the degrees are the same, there is no slant asymptote.

Caleb Alexander · Answer

undefined To find the asymptotes of the function $ g(x) = \frac{4x^2+11}{x^2+8x-9} $, we need to consider both vertical and horizontal asymptotes.

Vertical Asymptotes:
Vertical asymptotes occur where the denominator of the function becomes zero but the numerator doesn't. For this function, the denominator $ x^2 + 8x - 9 $ factors into $ (x - 1)(x + 9) $, so vertical asymptotes occur at $ x = -9 $ and $ x = 1 $.

Horizontal Asymptote:
To find the horizontal asymptote, we examine the behavior of the function as $ x $ approaches positive or negative infinity. Since both the numerator and the denominator are quadratic polynomials, we compare the degrees of the leading terms. In this case, the degrees are equal, so we divide the leading coefficients. The horizontal asymptote is the ratio of the leading coefficients of the numerator and denominator, which is $ y = \frac{4}{1} = 4 $.

Therefore, the asymptotes for the function $ g(x) = \frac{4x^2+11}{x^2+8x-9} $ are:

- Vertical asymptotes: $ x = -9 $ and $ x = 1 $
- Horizontal asymptote: $ y = 4 $