As you have said, just put the points into the equation:
\begin{align}
5 & = a^2 + b \\
-4 & = a^0 + b
\end{align}
The second equation directly gives you the first value you need, inserted into the first equation you can easily calculate $a$.
The curve you are trying to find an equation for also seems to pass through the point $(-5,15)$. Given three points, it's always possible to find a parabola that passes through them, that is, an equation of the form $y=ax^2+bx+c$. To find the coefficients $a$, $b$, and $c$, you need to plug in the $x$ and $y$ coordinates of the three points, $(5,17)$, $(-5,15)$, and $(-9,13)$:
$$\begin{align}
17&=25a+5b+c\\
15&=25a-5b+c\\
13&=81a-9b+c\\
\end{align}$$
When I solve this, I get the somewhat ungainly quadratic equation
$$y=-{3\over140}x^2+{1\over5}x+{463\over28}$$
One test of the appropriateness of this solution is how close its curve passes to the concentric circles; if it intersects any of them, then a parabola is not what you want.
It would help to know something about the context of this problem. Is it an assignment for a math class in which you are learning about logarithms and graphing equations, or is it part of some graphic arts course that assumes you already have facility with equations? The top figure clearly indicates one curve is logarithmic, but it offers no hint for the curve you're asking about. There are lots and lots of mathematical ways to get a curve that has the look you're looking for; the question is, what sort of math are you expected to or willing to use?
Best Answer
After determining the asymptote of your exponential equation, you have two unknowns but only one equation. Your system of equations is thus underdetermined and has infinitely many solutions. For example, the graphs of $4\cdot2^{-x}-2$ and $8\cdot2^{-2x}-2$ have their horizontal asymptote at $y=-2$ and pass through $(1,0)$, but are different.
To find a specific solution you need to fix one of the unknowns arbitrarily and solve for the other. In this case, we know that $0=Ae^{3B\ln2}-2$, or $2=A8^B$. If we set $A=1$ we find that $B=\frac13$; if we set $B=1$ we find that $A=\frac14$. Hence $y=e^{\frac x3}-2$ and $y=\frac14e^x-2$ are possible equations for the graph.
If you had been given one more point on the graph as you wished, like the $y$-intercept, you would be able to find a unique equation.