Constant rate of change for $f(x) = -x^2 +7$ I'm taking Calculus I and we are going over Average Rate of Change and the Difference Quotient in preparation to discuss derivatives. 
The current investigation states:
Let $f(x) = -x^2 + 7$.
The question asked if $f(x)$ varies at a constant rate of change, which I decided it does not, since it is exponential and thus has some acceleration/deceleration. What I'm having trouble grasping is the next part. It asks

What is the constant rate of change of the linear function $g$ that has the same change in output values over the interval $x=-3$ to $x=5$ as the function $f$?

I was searching for resources trying to help with this problem and found How are the average rate of change and the instantaneous rate of change related for ƒ(x) = 2x + 5?
So I applied the idea to my problem:
$$\frac{(b^2+7)-(a^2+7)}{b-a} = \frac{b^2-a^2}{b-a}=\frac{(b+a)(b-a)}{b-a}=b+a$$ 
But I didn't really know how to use this afterwards, since in the answer there, they weren't left with the variables they substituted in. How can I find the rate of change like this?

The relevant points on the graph of the function $f$ are $(-3, -2)$ and $(5, -18)$ (evaluate $f(-3)$ and $f(5)$ to determine this). A linear function $g$ passing through these two points would have slope (i.e., constant rate of change) equal to
$$
\frac{-18-(-2)}{5-(-3)} = -2.
$$
Your attempt at a general solution could be made to work (there was just an typo in your starting point). The slope of the line through the points $(a, f(a))$ and $(b, f(b))$ is given by
$$
\frac{f(b) - f(a)}{b-a} = \frac{(-b^2 + 7) - (-a^2 + 7)}{b-a} = \frac{a^2 - b^2}{b-a} = -(a + b),
$$
which gives $-2$ in our particular case.
