The reason why $\frac{a+b}{2}=c$ in a quadratic.

Question

Question: In a quadratic, I learned that there's a point $c$ on the graph that has the same slope (or derivative?) as the slope that any two points $a$ and $b$ on the graph form, and that is expressed as $$\frac{f(b)-f(a)}{b-a}=f'(c).$$ And, I learned that $\frac{a+b}{2}=c$. My tutor said he will show me the proof when he comes back, but I'm curious. Can somebody prove that thing?.

Answer 1

If the result is true, then you can prove it by direct verification:

Guide:

• Let the quadratic function be $f(x) = \alpha x^2 + \beta x + \gamma$
• Compute $f'(x)$.

• Compute $f'(c)= f'\left(\frac{a+b}2\right)$ in terms of $a$ and $b$.

• Compute $f(b)$, $f(a)$ and then $\frac{f(b)-f(a)}{b-a}$, verify that it is equal to the expression that you obtained above.

Note: $$f(b)-f(a)=\alpha(b^2-a^2)+\beta(b-a)$$

The following identities might help you:

$$b^2-a^2=(b-a)(b+a)$$

Edit after OP attempted:

\begin{align}\frac{f(b)-f(a)}{b-a} &=\frac{\alpha(b^2-a^2)+\beta(b-a)}{b-a}\\ &=\alpha(b+a)+\beta\\ &= 2\alpha \left( \frac{a+b}2\right)+\beta \\ &= 2\alpha(c)+\beta \\ &= f'(c)\end{align}