This is how I would attempt this
Ok first for simplicity lets call:
the line made by joining points $(-17,0)$ and $(-3,9)$ as Line A
and the line made by joining points $(-12,0)$ and $(-3,9)$ as Line B
and the left curve and right curve as shown in the diagram.
Now to find the line of A and B is pretty simple, Il do for one (Line A) to demonstrate it.
The equation of a striaght line is $y=mx+c$
where m is the gradient and c is the intercept.
$m=\frac{y_1-y_2}{x_1-x_2}$
where in Line A the values are :
$x_1=-17$
$x_2=-3$
$y_1=0$
$y_2=9$
Now plugin these and then to now since you know the value of m, to find c
you have
$y=mx+c$
$c=y-mx$
Put any co ordinate out of the two. Il put $(-3,9)$
and you get
$c=9-m(-3)$
Because you know m you can find c, and this is the equation of line A. Follow the method for line B.
Now for the curves, lets first look at curve on the right side
We know three points on the curve which are $(-27,17)$ , $(-9,3)$ and $(-3,9)$
and we know its in the form $y=a+c log_b(dx)$
But there are 4 unkowns you need to solve for a,b,c,d. So that means you need 4 equations. So the first three equations will be subsituting the 3 points that they have given into the equation. Which gives:
$17=a+c log_b(-27d)$
$3=a+c log_b(-9d)$
$9=a+c log_b(-3d)$
And in order to get the 4th equation you should know that, the line A is normal to the point (-3,9) on the curve, this means that you need to differentiate the function $y=a+c log_b(dx)$ and then you get the first derivative which is a function.
Then to this when you substitute x=-3 to this function ($dy/dx$) you get an expression with a bunch of unknowns. Lets call this expression as T. NOW we want a formula (not an expression) so we should know that the gradient of line A (which you should know by now) multiplied by the T is equal to -1. (This is a rule)
So there you have it a 4th equation, now you can solve the 4 equations, find the four unknowns and plot this curve.
Now the curve at the Right side , it approximately looks like the mirror image of the curve on the left side at a vertical line (which I will tell in a moment) and they have mentioned a point it meets, so I would say the right hand side is the reflection of the curve on the left side about the line $x=-9$ (the line of reflection). Look closely, or better yet draw the line $x=-9$ and this will become clear to you.
So since you know its the reflection of the left curve (a known curve by now) and you know a point it meets, and you know the line of reflection. I dont think it would be hard to proceed from this point on. Try these, if you get stuck ;) Dont forget to ask.
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
The original picture tells you the curve has the form $y=a+c\log_b(dx)$ for some constants $a$, $b$, $c$, and $d$, and that it passes through the points $(-27,17)$, $(-9,13)$, and $(-3,9)$. Since $27$, $9$, and $3$ are powers of $3$, it makes sense to try $b=3$ for the base of the logarithm (as you did). If you let $d=-1$ (it has to be negative because your $x$'s are negative and $\log(dx)$ is defined only when $dx$ is positive), then you get
$$\begin{align} 17&=a+3c\\ 13&=a+2c\\ 9&=a+c\\ \end{align}$$
If you pick any two of these equations and solve for $a$ and $c$, you get $a=5$ and $c=4$. (For example, subtracting the second equation from the first gives $17-13=(a+3c)-(a+2c)$, which reduces to $4=c$. Note, there are three equations here in only two unknowns, so you have to check that the answer really does satisfy all three equations, but of course it does.)