Question

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

Explanation:

Step1: Factor Denominator

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

Step2: Find Undefined Points

Set denominator equal to zero: \(4(x - 2)(x + 1)=0\). Solve for \(x\): \(x - 2 = 0\) or \(x + 1 = 0\), so \(x = 2\) or \(x=-1\).

Step3: Check Numerator at These Points

Numerator at \(x = 2\): \(2^2 + 4 = 8
eq0\). Numerator at \(x=-1\): \((-1)^2 + 4 = 5
eq0\). So vertical asymptotes at \(x = 2\) and \(x=-1\).

Answer:

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