Question

consider the following equation:
$f(x) = \frac{x^2 + 4}{4x^2 - 4x - 8}$
name the vertical asymptote(s).

Explanation:

Step1: Factor the denominator

Factor \(4x^{2}-4x - 8\). First, factor out a common factor of \(4\): \(4(x^{2}-x - 2)\). Then, factor the quadratic \(x^{2}-x - 2=(x - 2)(x+1)\). So the denominator is \(4(x - 2)(x + 1)\).

Step2: Set the denominator equal to zero

Set \(4(x - 2)(x + 1)=0\). Using the zero - product property \(ab = 0\) implies \(a = 0\) or \(b = 0\). Here, \(x-2=0\) gives \(x = 2\) and \(x + 1=0\) gives \(x=-1\).

Step3: Check for common factors with the numerator

The numerator \(x^{2}+4=(x + 2i)(x - 2i)\) (using complex numbers, but in the real - number system, it cannot be factored). Since there are no common factors between the numerator \(x^{2}+4\) and the denominator \(4(x - 2)(x + 1)\) in the real - number system.

Answer:

The vertical asymptotes are \(x=-1\) and \(x = 2\)