How do you find the vertical, horizontal or slant asymptotes for #f(x) = ((x-3)(9x+4))/(x^2-4)#?

Question

Anna Allen · Accepted Answer

The vertical asymptotes are #x=2# and #x=-2#
The horizontal asymptote is #y=9# The denominator, #x^2-4=(x+2)(x-2)# As we cannot divide by zero, the vertical asymptotes are #x=2# and #x=-2# As the degree of the numerator is the same asthe degree of the denominator, there is no slant asymptote: #lim_(x->+-oo)f(x)=lim_(x->+-oo)(9x^2)/x^2=9# #:. y=9# is a horizontal asymptote graph{(y-((x-3)(9x+4))/((x+2)(x-2)))(y-9)=0 [-41.13, 41.15, -20.55, 20.57]}

Elijah Allen · Answer

undefined To find the vertical, horizontal, or slant asymptotes for $ f(x) = \frac{(x-3)(9x+4)}{x^2-4} $, we first need to simplify the function and then analyze its behavior as $ x $ approaches infinity or certain values.

1. **Vertical Asymptotes**: Vertical asymptotes occur where the denominator of a rational function is zero but the numerator is not zero. In this case, the denominator $ x^2 - 4 $ is zero when $ x = -2 $ or $ x = 2 $. However, both $ x - 3 $ and $ 9x + 4 $ are not zero at these points, so $ x = -2 $ and $ x = 2 $ are vertical asymptotes.

2. **Horizontal Asymptotes**: Horizontal asymptotes occur when $ x $ approaches infinity. To find them, we compare the degrees of the numerator and denominator. Since both have degree 2, we divide the leading coefficients (the coefficients of the highest-degree terms). The leading coefficient of the numerator is 9, and the leading coefficient of the denominator is 1. Therefore, the horizontal asymptote is $ y = 9/1 = 9 $.

3. **Slant Asymptotes**: Slant asymptotes occur when the degree of the numerator is exactly one more than the degree of the denominator. In this case, since the degrees are equal, there are no slant asymptotes.

In summary, for $ f(x) = \frac{(x-3)(9x+4)}{x^2-4} $:
- Vertical asymptotes are at $ x = -2 $ and $ x = 2 $.
- Horizontal asymptote is at $ y = 9 $.
- There are no slant asymptotes.