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Math Help - Describing end behavior

  1. #1
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    Describing end behavior

    f(x) = (3-x)/(x^2-16)

    Describe the behavior of f(x) to the left and right of each asymptote.
    Asymptotes are y= 4 and y= -4

    The right behavior of the first asymptote at y = 4 is ∞correct? Would the left behavior at asymptote then be - ∞?

    And the right behavior of the asymptote y = -4 would also be ∞ and the left would again be -∞?

    Would someone mind correcting this if necessary?
    http://www.walterzorn.com/grapher/grapher_e.htm -here is a grapher if you need.

    Thanks so much!
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  2. #2
    Senior Member pacman's Avatar
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    murphie, this graph may help you visualize the graph .. . .

    Attached Thumbnails Attached Thumbnails Describing end behavior-sert.gif  
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  3. #3
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    I'm confused, the graph from my calculator looks completely different, your graph suggests that the left behavior at y=-4 is - infinity, and - infinity on the right?? and for the asymptote at y=4 the left is - infinity and right is infinity?
    I am fairly bad at calculus so feel free to correct me if i said something stupid. But thanks for your help.
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  4. #4
    Senior Member pacman's Avatar
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    your equation can be re-arranged into this,

    y = (3 - x)/((x - 4)(x + 4)), that means x = -4 and x = 4 are vertical asymptotes.

    y approaches zero as x approaches positive or negative infinity
    Attached Thumbnails Attached Thumbnails Describing end behavior-fgh.gif   Describing end behavior-qwe.gif  
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