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.

option a.

step-by-step explanation:

if you look closely at the graph, we can notice that the function is not defined at two different points.

if you look at functions b and c, you will notice that both functions are not defined at just one point. the function b is not defined at x= -5/2 and function c is not defined at x=-5. both options are discarded, given that our function should not be defined at two fiferent points.

now, if you look closely at the graph you will notice that the function sketched is not defined at x=5 and x=-2.

really really sorry if i'm

-12 -5(x + 4) = 3x

-12 - 5x - 20 = 3x

-12 - 20 - 5x = 3x

-32 - 5x = 3x

    + 5x   + 5x

-32 = 8x

8         8

- 4= x

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