Calculus one help?

I need help finding the vertical and horizontal asymptotes of this function. I'm mostly stuck with the vertical asymptotes but I would appreciate help with both!! 

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  • 1 month ago

    g(x) = 2x³/(x³ + 8)

    Dividing through by x³ we get:

    g(x) = 2/(1 + 8/x³)

    As x --> ±∞, 8/x³ --> 0

    so, g(x) --> 2/(1 + 0)...i.e. 2

    Hence, g(x) = 2 is a horizontal asymptote

    Vertical asymptotes occur when g(x) does not exist

    i.e. when x³ + 8 = 0

    so, when x = -2...vertical asymptote

    A sketch is below.

    :)> 

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  • 1 month ago

    Remember that (x³+8)=(x+2))(x²-2x+4) and g(-2)=∄ because (x+2)=0 in the denominator

    So your answer is x=-2⁻ and x=-2⁺

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