Question

find the horizontal and vertical asymptotes for the function.
$f(x) = \frac{4x^2 + 2x - 12}{x^2 - 4}$

Explanation:

Step1: Find Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero (and numerator is not zero). Factor denominator: \(x^2 - 4=(x - 2)(x + 2)\). Set denominator to zero: \((x - 2)(x + 2)=0\Rightarrow x = 2\) or \(x=-2\). Check numerator at these points:

• At \(x = 2\): \(4(2)^2+2(2)-12=16 + 4-12 = 8

eq0\)

• At \(x=-2\): \(4(-2)^2+2(-2)-12=16-4 - 12 = 0\). So \(x=-2\) is a hole (since numerator and denominator both zero), not an asymptote. Thus, vertical asymptote at \(x = 2\).

Step2: Find Horizontal Asymptotes

For rational functions \(\frac{ax^n+...}{bx^m+...}\), if \(n = m\), horizontal asymptote is \(y=\frac{a}{b}\). Here, \(n = m = 2\), \(a = 4\), \(b = 1\). So horizontal asymptote \(y = 4\).

Answer:

Vertical Asymptote: \(x = 2\)
Horizontal Asymptote: \(y = 4\)