Question

g rational functions
ying asymptotes
this is a multi - part item.
consider the following equation:
$f(x)=\frac{x^{2}+4}{4x^{2}-4x - 8}$
name the vertical asymptote(s).
1 $x = - 1$ and $x = 2$
because
$mlt n$
$m = n$
$a_{m}lt b_{n}$
$a_{m}=b_{n}$
this is where the function is
undefined

Explanation:

Step1: Recall Vertical Asymptote Rule

Vertical asymptotes of a rational function \( f(x)=\frac{N(x)}{D(x)} \) occur where \( D(x) = 0 \) (and \( N(x)
eq0 \) at those points, to avoid holes).

Step2: Analyze Denominator \( D(x)=4x^2 - 4x - 8 \)

Factor denominator: First, factor out 4: \( 4(x^2 - x - 2) \). Then factor quadratic: \( x^2 - x - 2=(x - 2)(x + 1) \). So \( D(x)=4(x - 2)(x + 1) \). Set \( D(x)=0 \): \( 4(x - 2)(x + 1)=0 \) gives \( x = 2 \) or \( x=-1 \). Check numerator \( N(x)=x^2 + 4 \): at \( x = 2 \), \( N(2)=4 + 4 = 8
eq0 \); at \( x=-1 \), \( N(-1)=1 + 4 = 5
eq0 \). So vertical asymptotes at \( x=-1 \) and \( x = 2 \) because these are the \( x \)-values where the function is undefined (denominator zero, numerator non - zero).

Step3: Match with Given Options

The reason for vertical asymptotes is that they occur where the function is undefined (denominator zero, numerator non - zero). So the correct option is "this is where the function is undefined".

Answer:

this is where the function is undefined