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Math Help - Horizontal asymptote

  1. #1
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    Post Horizontal asymptote

    Hi,
    I have just started precal and I can't understand well how to find the horizontal asymptote for f(x)=(4|x-2|)/x+1. Can somebody help me. Thank you.
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  2. #2
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    Quote Originally Posted by cuteisa89 View Post
    Hi,
    I have just started precal and I can't understand well how to find the horizontal asymptote for f(x)=(4|x-2|)/x+1. Can somebody help me. Thank you.
    You need to think about what happens when x goes to + infty and also
    to -infty.

    For large x:

    f(x) = (4x-8)/(x+1)

    divide top and bottom by x:

    f(x) = (4-8/x)/(1+1/x)

    so as x -> infty, f(x)->4 (as 8/x and 1/x both go to zero). So f(x)=4 (or y=4)
    is an asymtot for large x.

    For large negative x, and large |x|:

    f(x) = -(4x-8)/(x+1)

    divide top and bottom by x:

    f(x) = (-4+8/x)/(1+1/x)

    so as x -> infty, f(x)->-4 (as 8/x and 1/x both go to zero). So f(x)=-4 (or y=4)
    is an asymtot as x->-infty.
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