Is the average of two endpoint's instantaneous rates of change always the same as the average rate of change of the interval on a parabola? Is the average of two endpoint's instantaneous rates of change always the same as the average rate of change of the interval on a parabola? If not, which additional circumstances would cause it to always be the same?
EDIT: How would you verify this algebraically? Answered
EDIT 2: They both result in the same answer, but do they get their answer from the same line that touches/intercepts the graph, or do they get their answer from different lines (secant, tangent, etc.) that touch/intercept the graph, and because the two lines are always parallel to each other, they result in the same answer?
 A: I'll give you a hint, but you need to know some calculus to understand the problem itself.
Recall that average value of a function $f$ on the interval $[a,b]$ is: $$\frac{f(b)-f(a)}{b-a}$$
And the average of the two endpoints instantaneous rate of change is given by:
$$\frac{f'(b) + f'(a)}{2}$$
Where $f'(x)$ is the derivative at some point $x$.
Now think about a polynomial of the form $f(x) = x^2$. How would you go about showing the two expressions above are the same? And then you could do the same kind of algebraic manipulation for any parabola of the form $f(x) = a(x-h)^2 + k$ with $a,h,k$ some constants. It's just more algebra.
A: A parabola is the graph of a quadratic polynomial $f(x)=ax^2+bx+c.$ The average rate of change of $f$ on the interval $[r,s]$ is $\frac{f(s)-f(r)}{s-r}.$ You can check that this computes to $\frac{a(s^2-r^2)+b(s-r)}{s-r} = a(s+r)+b.$ The instantaneous rate of change of a function is its derivative $f'$. In this case $f'(x)=2ax+b,$ so the instantaneous rates at the endpoints are $f'(s)=2as+b$ and $f'(r)=2ar+b.$ The average of these two values is easily seen to be $a(s+r)+b,$ which is what you wanted.
