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    http://www.examsolutions.net/a-level...e/paper.php#Q5

    part c i don't understand why the range is just -k

    usually i let x=0 and e to the 0 makes 1 so from my logic(which isn't right this time) i got 1-k
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    e^(2x) is always greater than zero (draw the graph)

    So e^(2x)-k is always greater than -k.
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    (Original post by thegreatwhale)
    http://www.examsolutions.net/a-level...e/paper.php#Q5

    part c i don't understand why the range is just -k

    usually i let x=0 and e to the 0 makes 1 so from my logic(which isn't right this time) i got 1-k
    For the graph f(x) = e^x, you know there is an asymptote at f(x) = 0.

    If the graph is e^2x, the asymptote will not move.

    The -k means all y values go downwards by k, and hence, since y must be greater than 0, it now must be greater than -k.
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    (Original post by Math12345)
    e^(2x) is always greater than zero (draw the graph)

    So e^(2x)-k is always greater than -k.
    (Original post by JLegion)
    For the graph f(x) = e^x, you know there is an asymptote at f(x) = 0.

    If the graph is e^2x, the asymptote will not move.

    The -k means all y values go downwards by k, and hence, since y must be greater than 0, it now must be greater than -k.
    ah thanks all
 
 
 
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