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

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 4: $4(x^2 - x - 2)$
Then factor the quadratic: $4(x-2)(x+1)$

Step2: Find denominator roots

Set denominator equal to 0:
$4(x-2)(x+1)=0$
Solve for $x$:
$x-2=0 \implies x=2$
$x+1=0 \implies x=-1$

Step3: Check numerator at roots

Numerator $x^2+4$ at $x=2$: $2^2+4=8
eq0$
Numerator at $x=-1$: $(-1)^2+4=5
eq0$
No common factors, so these are vertical asymptotes.

Answer:

The vertical asymptotes are $x=-1$ and $x=2$