Rates of Change in Linear and Quadratic Functions

$\frac{f(3) - f(1)}{3 - 1}$$= \frac{13 - 3}{2}$$= 5$, the slope of the secant line through $(1, 3)$ and $(3, 13)$.

• Answer: 5

Over steps of $1$, the first differences are the average rates of change on $[0, 1], [1, 2], \ldots$. They change by the same amount, $4$, each time: the rates change linearly, which is what a quadratic does.

• The values fit a linear function, because the change in the change is the same at each step of the table.: A linear function has constant first differences; these first differences, $-1, 3, 7, 11, 15$, are not constant.
• Correct: The values fit a quadratic: the rates on consecutive unit intervals rise by $4$ each step.: Right: constant second differences over equal steps are the mark of a quadratic.
• $f$ is increasing on $[0, 5]$.: $f(1) = 3 < f(0) = 4$, so $f$ decreases on $[0, 1]$.
• The average rate of change of $f$ is $4$ on each interval of length $1$.: $4$ is the change in the rates, not a rate; the rates are $-1, 3, 7, 11, 15$.

Over equal steps, increasing average rates of change mean the rate of change is increasing, so $f$ is concave up.

• It is concave down, because the second differences are positive numbers that stay the same.: Positive second differences mean the rates are going up, which is concave up.
• It is concave up, because $f(5) > f(0)$.: Comparing two outputs gives direction, not concavity. Concavity comes from how the rates change.
• Correct: It is concave up, because the average rates over consecutive equal intervals increase.: Right: the rates $-1, 3, 7, 11, 15$ increase.
• It is a straight line, because the second differences are constant.: A line has constant first differences; constant second differences mean a parabola.

$h(0) = 1$ and $h(3) = -45 + 60 + 1 = 16$. Slope $= \frac{16 - 1}{3 - 0} = 5$ meters per second, the average rate of change of $h$ on $[0, 3]$.

• Answer: 5 meters per second

The estimate is a point on the secant line: $y = 1 + 5(1) = 6$. $h$ is concave down on $[0, 3]$ (its rates fall), so its graph lies above the secant line between $t = 0$ and $t = 3$, and $6$ is an underestimate. Indeed $h(1) = 16$.

• $6$; an overestimate, since $h$ is increasing on $[0, 1]$: Whether $h$ rises does not decide over or under; the bending does. The graph is above the secant line.
• Correct: $6$; an underestimate, since $h$ is concave down on $[0, 3]$: Right: the secant line gives $1 + 5(1) = 6$, and the true height $h(1) = 16$ is above it.
• $16$; exact, since the point $(1, 16)$ is on the graph of $h$: $16$ is $h(1)$ itself; the secant line's value at $t = 1$ is $6$.
• $5$; an underestimate, because the slope is positive: $5$ is the slope. The estimate is the line's value at $t = 1$: $1 + 5 \cdot 1 = 6$.

$h(0),$$h(1),$$\ldots,$$h(4)$$= 1,$$16,$$21,$$16,$$1$, so the rates over unit intervals are $15, 5, -5, -15$. They change linearly (by $-10$ each step), so $h$ is concave down.

• Correct: They are $15, 5, -5, -15$: they decrease by $10$ each second.: Right: a quadratic's rates over consecutive equal intervals change by a constant amount.
• They are $16, 21, 16, 1$: they rise, then fall.: These are heights, not rates of change.
• They each equal $5$, the slope of the secant line on $[0, 3]$.: Only a line has the same rate on every interval.
• They are $15, 5, -5, -15$, which drop by $5$ from one to the next.: $15 - 5 = 10$: the rates drop by $10$ each step.