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$f(x)=A2^{kx}+B$

Is is possible to translate this function to the the following criteria:

$F(x)$ is always decreasing, has a horizontal asymptote at $y=1$, and goes through the point $(0,4)$?

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    $\begingroup$ Use 2^{kx}, with curly braces, to get the whole $kx$ into the exponent. $\endgroup$
    – Brian M. Scott
    May 8, 2012 at 5:48
  • $\begingroup$ Thanks, I was not sure how to do that. $\endgroup$
    – Kurt
    May 8, 2012 at 5:53
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    $\begingroup$ Are you trying to ask whether you can choose $A$, $B$ and $k$ to create a function which meets the criteria you have listed? In which case, since it is homework, you should have some idea how to express those criteria in ways which help you to relate them to the question. I think you need to tell us more about what you already know, and show some attempt to make progress - otherwise you'll get "the answer", but learn no maths or problem solving skills. $\endgroup$
    – Mark Bennet
    May 8, 2012 at 5:54
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    $\begingroup$ Ok, so from my understanding, this function is of the exponential variety. From my knowledge of translations I know that I could make is decreasing by changing it to $-2^x+1$, and that I could make it pass through the point (0,4) but adding 5 to it. But I am confused as to how it can be manipulated to not only pass through the point (0,4), but also have a horizontal asymptote at y=1. $\endgroup$
    – Kurt
    May 8, 2012 at 6:12
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    $\begingroup$ (+1) for writing out your thoughts. In the future, you might consider adding those to the body of the question, rather than leaving them in comments. $\endgroup$
    – The Chaz 2.0
    May 9, 2012 at 14:41

2 Answers 2

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To make the function pass through $\langle 0,4\rangle$, you need to choose $A$ and $B$ so that $f(0)=4$, i.e., so that $A2^0+B=4$. You already know what $B$ has to be in order to get the right horizontal asymptote, so there's only one choice for $A$; does it work, or does the resulting function fail to meet one of the criteria?

If you're allowed to choose $k$ as well as $A$ and $B$, note that it makes a difference whether $k$ is positive or negative.

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  • $\begingroup$ If $A=-1$ and $B=5$ $F(0)$ would satisfy the equation. But that would leave me horizontal Asymptote at 6. The resulting function fails to meet the criteria. $\endgroup$
    – Kurt
    May 8, 2012 at 6:34
  • $\begingroup$ @Kurt: But then you'd have your horizontal asymptote at $y=5$. You already picked the only possible value of $B$ to get $y=1$ as horiz. asymptote. $\endgroup$
    – Brian M. Scott
    May 8, 2012 at 6:36
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In order to find the correct $A$, $B$, and $k$, we have to determine what these values do to the function $f(x) = 2^x$. $f(x)$ is an exponential function that is always increasing, has a horizontal asymptote, H.A., at $y = 0$, and passes through the point $(0, 1)$. Now, $A$ will vertically stretch the graph of $f(x)$, $B$ will vertically shift the graph, and $k$ will determine if the graph is always increasing or always decreasing (it's essentially a horizontal reflection of the graph across the $y$-axis).

To determine $A$ and $B$, we have to combine what they are doing. $B$ will move the H.A. up or down. Since we want the H.A. to be $y=1$, let $B = 1$. As mentioned before, $A$ will vertically stretch $2^x$, meaning that the $y$-intercept will change to $(0, A)$. With this in mind, let $A = 3$. The function $3 \cdot 2^x + 1$ will have a H.A at $y = 1$ and it will pass through the point $(0, 4)$.

Since we want the function to always be decreasing, let $k = -1$.

Your required function will be: $F(x) = 3 \cdot 2^{-x} + 1$.

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