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Which statement describes whether the function f (x) = StartFraction 8 (x minus 1) squared Over (x minus 8) squared EndFraction has a vertical asymptote at x = 8? There is a vertical asymptote at x = 8 because Limit of f (x) as x approaches infinity = 8 There is not a vertical asymptote at x = 8 because Limit of f (x) as x approaches 8 does not exist. There is a vertical asymptote at x = 8 because Limit of f (x) as x approaches 8 minus = infinity and Limit of f (x) as x approaches 8 plus = infinity. There is not a vertical asymptote at x = 8 because Limit of f (x) as x approaches 8 minus and Limit of f (x) as x approaches 8 plus are not opposites.

Respuesta :

Using the vertical asymptote concept, it is found that there is not a vertical asymptote at x = 8 because [tex]\lim_{x \rightarrow 8^-} f(x)[/tex] and [tex]\lim_{x \rightarrow 8^+} f(x)[/tex]are not opposites.

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A vertical asymptote of a function happens at a point outside the function domain.

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In this question, the function is:

[tex]f(x) = \frac{8(x - 8)^2}{(x - 8)^2}[/tex]

In a fraction, the denominator cannot be 0, which could lead to a vertical asymptote at x = 8. However, the term with (x - 8) in the denominator can be factored out of the expression, and thus, an equivalent function is:

[tex]f(x) = 8[/tex]

And the limit at x = 8 will be:

[tex]\lim_{x \rightarrow 8^-} f(x) = \lim_{x \rightarrow 8^+} f(x) = 8[/tex]

Not opposite, so the correct option is:

There is not a vertical asymptote at x = 8 because [tex]\lim_{x \rightarrow 8^-} f(x)[/tex] and [tex]\lim_{x \rightarrow 8^+} f(x)[/tex]are not opposites.

A similar question is given at https://brainly.com/question/23535769

Answer:

C

Step-by-step explanation:

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