Question

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

name the vertical asymptote(s).
☑️ x = -1 and x = 2☑️
complete

because
○ $m < n$
○ $m = n$
○ $a_m < b_n$
○ $a_m = b_n$
○ this is where the function is undefined
done ☑️

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 \(x^{2}-x - 2=(x + 1)(x - 2)\). So the denominator is \(4(x + 1)(x - 2)\).

Step2: Find where the denominator is zero

Set the denominator equal to zero: \(4(x + 1)(x - 2)=0\). Using the zero - product property \(x+1 = 0\) gives \(x=-1\) and \(x - 2=0\) gives \(x = 2\). Vertical asymptotes occur at the values of \(x\) that make the denominator zero (provided the numerator is not also zero at those points). The numerator \(x^{2}+4=(x + 2i)(x - 2i)\) has no real roots.

Answer:

Vertical asymptotes are at \(x=-1\) and \(x = 2\) because this is where the function is undefined.