Find the slope of the chord joining two points on a quadratic

Let $f$ be the quadratic function

$f(x) = x^2 + ax + b.$

Find the slope of the chord joining the points on the graph of $f$ for $x = x_1$ and $x = x_2$ .
Find (in terms of $x_1$ and $x_2$ ) the values of $x$ for which the tangent at $(x,f(x))$ has the same slope as the chord in part (a).

The points on the graph of $f$ at $x_1$ and $x_2$ are $(x_1, f(x_1))$ and $(x_2, f(x_2))$ . So the chord joining them has slope given by the difference quotient (or “rise over run” if you like),
$\begin{align*} \frac{f(x_1) - f(x_2)}{x_1 - x_2} &= \frac{x_1^2 + ax_1 + b - x_2^2 - ax_2 - b}{x_1 - x_2} \\ &= \frac{(x_1^2 - x_2^2)}{x_1-x_2} + \frac{a (x_1 - x_2)}{x_1 - x_2} \\ &= x_1 + x_2 + a. \end{align*}$
The slope of the tangent line to the graph of $f$ is given by
$f'(x) = 2x + a.$

Since we calculate the slope of the chord in part (a) as $x_1 + x_2 + a$ we set these equal and obtain,

$2x+a = x_1 + x_2 + a \quad \implies \quad x = \frac{1}{2} (x_1 + x_2).$