Linear and Quadratic Rates — AP® Precalculus Notes

Topic 1.2 measured how fast a function changes with average rates of change. This topic asks how those rates behave for the two simplest families. Lines have one rate. Quadratics have rates that change at a steady pace. The difference shows up in tables, on graphs, and in the most-tested reasoning of the exam: estimating with a secant line and saying whether the estimate is too big or too small.

The secant line

A secant line is a line through two points of a graph. The secant line of $f$ through $(a, f(a))$ and $(b, f(b))$ has slope

$\frac{f(b) - f(a)}{b - a},$

which is exactly the average rate of change of $f$ on $[a, b]$ . Say it in one sentence: "The average rate of change of $f$ on $[a, b]$ is the slope of the secant line through $(a, f(a))$ and $(b, f(b))$ ." Graders look for that link.

Linear functions: one rate

For a linear function $g(x) = mx + b$ , every secant line is the line itself, so the average rate of change is $m$ on every interval. In a table at equal input steps, the first differences are all the same. For $g(x) = -3x + 7$ , each step of $1$ in $x$ changes $g$ by $-3$ .

Quadratic functions: rates that change linearly

For a quadratic, take equal input steps and list the changes in output (the first differences). Then list the changes in those (the second differences). The second differences are constant.

Worked example 1. $f(x) = 2x^2 - 3x + 4$ at $x = 0, 1, 2, 3, 4, 5$ .

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$4$	$3$	$6$	$13$	$24$	$39$
first differences		$-1$	$3$	$7$	$11$	$15$
second differences			$4$	$4$	$4$	$4$

Over steps of $1$ , each first difference is the average rate of change on that unit interval. The rates $-1, 3, 7, 11, 15$ go up by $4$ each time: they change linearly. That is what "quadratic" means for rates of change.

The pattern also predicts: the next first difference is $15 + 4 = 19$ , so $f(6) = 39 + 19 = 58$ .

The steps must be equal. If a table jumps from $x = 1$ to $x = 3$ , its differences are not comparable with the others until you divide by the step length.

Concavity from the rates

When the average rates over consecutive equal intervals increase, the rate of change is increasing and the graph is concave up. When they decrease, it is concave down. The table above has rates $-1, 3, 7, 11, 15$ : concave up. A ball's height $h(t) = -5t^2 + 20t + 1$ has rates $15, 5, -5, -15$ over the first four seconds: concave down.

For a quadratic $ax^2 + bx + c$ , the concavity is the same everywhere: concave up when $a > 0$ , concave down when $a < 0$ .

Estimating with a secant line

An estimate from a secant line uses a point on the secant line, not on the graph. Then concavity tells you how good it is.

Worked example 2. For $h(t) = -5t^2 + 20t + 1$ , use the secant line on $[0, 3]$ to estimate $h(1)$ , and say whether it is an overestimate or an underestimate.

$h(0) = 1$ , $h(3) = 16$ . Slope: $\frac{16 - 1}{3 - 0} = 5$ . Secant line: $y = 1 + 5t$ .

Estimate: $y = 1 + 5(1) = 6$ meters.

Reason: $h$ is concave down on $[0, 3]$ , so its graph lies above the secant line between $t = 0$ and $t = 3$ . The estimate is an underestimate. (Check: $h(1) = 16$ .)

The graph of f(x) = −0.5x² + 4x for 0 ≤ x ≤ 8 and the blue secant line through (1, 3.5) and (5, 7.5), which has slope 1. At x = 2 the secant line gives 4.5 (square) while the graph gives f(2) = 6 (dot): the graph is concave down and lies above the secant line between x = 1 and x = 5. — **Figure 1.** A concave down graph lies above its secant line: the secant value $4.5$ is below $f(2) = 6$ . LevlPrep original diagram.

On the interval, $f$ is	The graph is	A secant estimate is
concave up	below the secant line	an overestimate
concave down	above the secant line	an underestimate

Whether $f$ is increasing or decreasing does not matter here; the concavity on the stated interval does. Name the interval in your reason: " $h$ is concave down on $[0, 3]$ ".