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    Yusf's Avatar
    Yusf Posts: 198, Reputation: 3
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    #1

    Apr 12, 2016, 12:30 AM
    How will the asymptote be affected by this translation?
    I am struggling with number b where my logic is that the asymptote will not change. But the answers say I am wrong. Please help.
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    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Apr 12, 2016, 05:45 AM
    Think of the function 4f(x) as stretching the original function f(x) vertically by a factor of 4. All 'y' values become 4 times as large while the 'x' values remain the same. Thus the max point becomes (-2,20) - which I see is what you wrote on the paper. Similarly the asymptote becomes y=4.
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    Yusf Posts: 198, Reputation: 3
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    #3

    Apr 12, 2016, 07:09 AM
    But the equation y=1/(x+1) or any similar equation will have asymptote at y=0. And when y=4f(x), y=4/(x+1). Still asymptote at y equal to 0. Why is it different here?
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    ebaines Posts: 12,131, Reputation: 1307
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    #4

    Apr 12, 2016, 07:49 AM
    Quote Originally Posted by Yusf View Post
    But the equation y=1/(x+1) or any similar equation will have asymptote at y=0. And when y=4f(x), y=4/(x+1). Still asymptote at y equal to 0. Why is it different here?
    Because 4 times zero is zero, whereas 4 times 1 is 4. If the asymptote was at y=4, then the asymptote for y=4f(x) would be 4 x 4 = 16.
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    Yusf Posts: 198, Reputation: 3
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    #5

    Apr 13, 2016, 03:45 AM
    Right, thanks hahaha ☺

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