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Math Help - exponential equation transformation....

  1. #1
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    exponential equation transformation....

    Hi,

    I have the following problem that I am stuck on. I know it is probably something basic that I am missing and I would appreciate any and all help.

    Thanks,
    TC

    Write 2 different exponential equations with different bases if there is a y intercept of 5 and an asymptote at y=3.

    I have gotten this far....

    y = a^x + 4

    a is the base, I know that anything to the power of 0 equals 1 so this equation satisfies the y-intercept part but I am at a loss about finding some way to have an asymptote at y=3
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  2. #2
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    Hello, TC!

    This is trickier than you think . . .


    Write two different exponential equations with different bases
    which have a y-intercept of 5 and an asymptote at y=3.

    An exponential function can have a coefficient in front: . y \:=\:k\!\cdot\!a^x

    This equation has a horizontal asymptote at y = 0 (the x-axis).

    To have an asymptote at y = 3, we must "raise" the function 3 units.
    . . So we have: . y \:=\:k\!\cdot\!a^x + 3

    To have a y-intercept of 5, we want: . x = 0 and y = 5
    . . Substitute: . 5 \:=\:k\!\cdot\!a^0 + 3\quad\Rightarrow\quad k = 2

    Hence, the function is: . y \:=\:2\!\cdot\!a^x + 3<br />


    This equation has a graph with a y-intercept of 5 and an asymptote at y = 3.
    It already satisfies the problem . . . the base does not matter.

    So pick any two positive bases (other than 1).

    I'd offer: . \begin{Bmatrix}y \;= \;2\!\cdot\!10^x + 3 \\ y \;= \;2\!\cdot\!e^x + 3\end{Bmatrix}

    . . but you can use your imagination . . .


    Examples

    . . y \;=\;2\!\cdot\!\pi^x + 3

    . . y \;=\;2\!\cdot\left(\frac{1}{4}\right)^x + 3 \;=\;2\!\cdot\left(4^{-1}\right)^x + 3\quad\Rightarrow\quad y \;=\;2\!\cdot\!4^{-x} + 3

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  3. #3
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    Thank you so much! I was approaching it from the y-intercept not the asymptote.

    I appreciate the assistance.

    TC
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    we could also just change the x to a -x


    so two would be: y = 2 \cdot e^x + 3 \mbox { and } y = 2 \cdot e^{-x} + 3

    the asymptotes and intercepts would remain the same, but you would reflect one graph in the y-axis to get the other
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