Estimating a slope from data AB & BC

With no formula available, the only tool is the average rate of change over an interval, used as an approximation of the instantaneous rate. Every choice of interval gives a different estimate, and the interval that hugs the point most closely gives the best one.

Three patterns cover almost every table question. For a point $a$ with table neighbours on both sides:

1. Symmetric (centred): $f'(a) \approx \dfrac{f(a + h) - f(a - h)}{2h}$, using the nearest row on each side. Preferred whenever both sides are available.
2. Forward: $\dfrac{f(a+h) - f(a)}{h}$, using $a$ and the next row.
3. Backward: $\dfrac{f(a) - f(a-h)}{h}$, using the previous row and $a$.

When $a$ sits at the edge of the table, only a one-sided estimate exists and that is fine. When it does not, reaching past a nearer row to a further one throws away accuracy for nothing.

The rows need not be evenly spaced. The rule is unchanged: change in output over change in input, using the closest row on each side.

A liquid's temperature $H$, in degrees Celsius, is recorded at $t$ minutes:

1. $t = 0,\ 3,\ 5,\ 8,\ 12$
2. $H = 60,\ 68,\ 75,\ 84,\ 93$

To estimate $H'(5)$, the nearest rows on either side are $t = 3$ and $t = 8$:

$$H'(5) \approx \frac{H(8) - H(3)}{8 - 3} = \frac{84 - 68}{5} = 3.2 \text{ } ^\circ\text{C per minute}.$$

Compare the alternatives. A backward estimate on $[3, 5]$ gives $3.5$; a forward estimate on $[5, 8]$ gives $3$; the whole table gives $\frac{93 - 60}{12} = 2.75$. The first two are defensible one-sided estimates and the last one is not, because it averages away eleven of the twelve minutes that were not being asked about.

Every one of these numbers is an average rate over an interval. It is being used to approximate an instantaneous rate, and the two are never claimed to be equal.

Given a printed graph, the procedure is to draw the tangent at the point by eye, then find its slope. The slope is read from two points that lie on the tangent line, chosen where it crosses grid intersections so the coordinates can be read exactly.

Two points on the curve give a secant slope, which is the wrong object once a tangent has been drawn. The tangent may cross the curve elsewhere, and that crossing point is not a legal reference either.

Reading the height of the curve instead of the slope is the other frequent slip. The question asks how steep, not how high, so the answer is a ratio of two grid counts and never a single coordinate.

A quick sanity check costs nothing: a rising graph must produce a positive number, a falling one a negative number, and a flat spot must produce something near zero.

This is one of the topics where a calculator earns its place. A numerical derivative feature evaluates a symmetric difference quotient at a very small $h$, which is exactly the procedure above with the interval driven down until the display stops changing.

Three habits keep the estimate meaningful:

1. Report the units. Output units per input unit, always. A bare number answers half the question.
2. Say that it is an estimate. On the AP exam the language matters: an approximation from a table is described as such, not written as an equality.
3. Check the sign and size against the picture. If the data are rising steadily and your estimate is negative, the subtraction ran backwards somewhere.

Concavity gives one more piece of information for free. If $f$ is increasing and concave up, then $f'$ is increasing, so a forward difference on $[a, a+h]$ comes out above $f'(a)$ and a backward difference on $[a-h, a]$ comes out below it. Knowing which side of the truth an estimate falls on is often worth as much as the estimate.