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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?

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    $f(x)$ is quadratic ($f(x) = ax^2 + bx + c$) not exponential ($f(x) = ab^{cx}$) – 2017-02-09
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    I think the part in the yellow box is just a convoluted way of asking "What is the average rate of change of $f$ from $x = -3$ to $x = 5$?". – 2017-02-09
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    Hint: $f(-3)=-2\,$, $f(5)=-18\,$, $\left(f(5)-f(-3)\right) / (5+3) = \cdots\;$ – 2017-02-09

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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.