Exponential translation $f(x)=A2^{kx}+B$
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 $(0,4)$? 
 A: 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.
A: 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: $F(x) = 3 \cdot 2^{-x} + 1$.
