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16 September, 03:38

What are the vertical asymptote (s) and hole (s) for the graph of y =

X-1 / - x² + 6x + 8

asymptote: x = 1 and holes: x = - 4,-2

asymptotes: x = - 4,-2 and no holes

asymptotes: x = - 4,-2 and hole: x = 1

asymptote: x = 1 and no holes

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Answers (1)
  1. 16 September, 03:55
    0
    asymptotes: x = - 4,-2 and no holes

    Step-by-step explanation:

    First we need to factor the denominator of the function

    y = (x-1) / (x2 + 6x + 8)

    x2 + 6x + 8 = 0

    delta = b2 - 4ac = 36 - 32 = 4

    x1 = (-6 + 2) / 2 = - 2

    x2 = (-6 - 2) / 2 = - 4

    So we have that x2 + 6x + 8 = (x+4) (x+2)

    So our function is:

    y = (x-1) / (x+4) (x+2)

    As there is no common expressions in the numerator and denominator, there are no holes.

    The asymptotes are when the denominator is zero, so:

    (x+4) (x+2) = 0

    x = - 4 or x = - 2
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