Slope of a secant line — the first building block of derivatives
A secant line is a straight line that cuts through a curve at two points. Its slope is just "rise over run" between those two points — the average rate of change of the function between them.
The tangent line is the special case where the two points collapse to one — that's the derivative, which you'll meet in the next pretest.
For any function $f$ and two x-values $a$ and $b$, the slope of the secant line through $(a, f(a))$ and $(b, f(b))$ is:
$$m_{\text{sec}} \;=\; \frac{f(b) - f(a)}{b - a}$$Whenever a pretest asks for a secant slope, your first move is always: find the two y-values, then divide.
Given $f(x) = x^2$, with $a = -1$ and $b = 3$:
Step 1 — Evaluate $f$ at each x-value.
Step 2 — Plug into the secant formula.
$$m_{\text{sec}} \;=\; \frac{f(3) - f(-1)}{3 - (-1)} \;=\; \frac{9 - 1}{3 + 1} \;=\; \frac{8}{4} \;=\; 2$$So the secant line through $(-1, 1)$ and $(3, 9)$ has slope:
The secant slope is the finite-difference version of the derivative. The key connection between this pretest and the next unit:
The derivative of $f$ at a point $c$ is the limit of secant slopes as the two x-values squeeze in around $c$:$$f'(c) \;=\; \lim_{b \to c} \frac{f(b) - f(c)}{b - c}$$
For $f(x) = x^2$ you can check this: $f'(x) = 2x$, so $f'(2) = 4$. If you computed the secant slope between $x=2$ and $x=2+h$ and took $\lim_{h \to 0}$, you'd get 4. Same machinery, different limit.
Pretest check. If $g(x) = x^2 + 3x$ and you want the secant slope through $x = 0$ and $x = 4$, what's the answer?
Hint: $g(0) = 0$, $g(4) = 16 + 12 = 28$. Now find the slope.
$m_{\text{sec}} = \dfrac{g(4) - g(0)}{4 - 0} = \dfrac{28 - 0}{4} = \boxed{7}$
Two natural directions from here:
Send the next problem whenever you're ready, or just tell me which of the three directions to take.