To determine the vertical asymptotes of the function [tex]\( f(x) = \frac{x^2 - 64}{8x - 64} \)[/tex], we need to identify the values of [tex]\( x \)[/tex] that make the denominator zero since those values create undefined points in the function.
1. Start with the denominator:
[tex]\[
8x - 64
\][/tex]
2. Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[
8x - 64 = 0
\][/tex]
[tex]\[
8x = 64
\][/tex]
[tex]\[
x = 8
\][/tex]
3. We found [tex]\( x = 8 \)[/tex] makes the denominator zero. This suggests a potential vertical asymptote at [tex]\( x = 8 \)[/tex].
4. Next, we examine the numerator [tex]\( x^2 - 64 \)[/tex] to ensure there is no cancellation between the numerator and the denominator at [tex]\( x = 8 \)[/tex]:
[tex]\[
x^2 - 64 = (x - 8)(x + 8)
\][/tex]
5. We substitute [tex]\( x = 8 \)[/tex] into the numerator:
[tex]\[
8^2 - 64 = 64 - 64 = 0
\][/tex]
Since the numerator also becomes zero when [tex]\( x = 8 \)[/tex], this means the factor [tex]\((x - 8)\)[/tex] cancels with the [tex]\((8x - 64)\)[/tex] denominator term. However, even with the cancellation, [tex]\( x = 8 \)[/tex] does create a point where the function is undefined. Thus, the only vertical asymptote remains at [tex]\( x = 8 \)[/tex].
6. There are no other values that make the denominator zero and create additional vertical asymptotes.
Based on this detailed analysis, the correct statement is:
The graph has a vertical asymptote at [tex]\( x=8 \)[/tex] only.
