Lesson 02: Rates of Change — Linear & Quadratic Functions

Warm-Up

Two savings plans, both starting at $0. Plan L: deposit $50 every week. Plan Q: deposit $10 the first week, $20 the second, $30 the third, and so on.

Week-by-week balances: Plan L: 50, 100, 150, 200, … Plan Q: 10, 30, 60, 100, …

Plan L grows by the same amount each week. Plan Q grows by amounts that themselves grow by the same amount ($10 more each week). That's the entire difference between linear and quadratic functions, seen in a bank account. Today we make it precise.

Core Concept

Linear functions: constant rate of change

A linear function f(x) = mx + b has average rate of change m over every interval:

AROC over [a, a+h] = ( m(a+h) + b − (ma + b) ) / h = mh / h = m

Equivalently: over equal-length input intervals, the output changes by equal amounts. In a table with equally spaced x-values, the first differences (each output minus the previous) are constant.

Constant first differences of −3 with step 1 → linear with slope −3.

⚠️ If the x-step isn't 1, the slope is Δoutput / Δinput. Differences of −6 with x-step 2 still mean slope −3.

Quadratic functions: linearly changing rate of change

A quadratic f(x) = ax² + bx + c does not have one rate of change — but its rate of change changes in the most orderly way possible: linearly.

Over equal-length input intervals, the first differences of a quadratic change by a constant amount, so the second differences (differences of the differences) are constant.

For f(x) = x²:

The useful algebra fact (worth seeing once): over consecutive intervals of length h, a quadratic's second difference is always 2ah². So from a table you can even recover a: a = Δ² / (2h²).

The diagnosis routine

Given a table with equally spaced inputs:

1. First differences constant → linear
2. First differences not constant, second differences constant → quadratic
3. Neither → some other family (exponential functions — where ratios are constant — arrive in Unit 2)

This routine is the seed of model selection (Lesson 9 and FRQ 1): real data won't be perfectly constant, and you'll say "roughly constant second differences suggest a quadratic model."

Concavity and the leading coefficient

For f(x) = ax² + bx + c:

• a > 0: second differences positive → rates of change increasing → concave up everywhere → the vertex is a global minimum
• a < 0: rates decreasing → concave down everywhere → vertex is a global maximum

The vertex sits at x = −b/(2a); a parabola is symmetric about that vertical line. Two zeros, when they exist, sit symmetrically around it — so the vertex x-coordinate is also the average of the zeros.

[GRAPH: Two parabolas on one set of axes. Left: f(x) = x² − 4x + 3 opening upward, vertex (2, −1) marked as global minimum, zeros at x = 1 and x = 3, axis of symmetry x = 2 dashed. Right: g(x) = −x² − 2x + 3... shown shifted right for clarity, opening downward, vertex marked as global maximum. Annotation: "a > 0: concave up, min" and "a < 0: concave down, max".]

AROC over an interval of a quadratic — the midpoint fact

A tidy property the exam likes: the average rate of change of a quadratic over [a, b] equals the (instantaneous) rate of change at the midpoint (a+b)/2. You saw this accidentally in Lesson 1, Example 4: AROC of x² over [2.9, 3.1] was exactly 6 — the rate at the midpoint x = 3. You don't need calculus to use the consequence: for a quadratic, AROCs over intervals with the same midpoint are all equal.

Worked Examples

Example 1 (easy) — Classify from a table 🚫 No-Calc

Problem: Equally spaced inputs; classify each as linear, quadratic, or neither.

Table 1: outputs 5, 8, 11, 14, 17 Table 2: outputs 2, 3, 6, 11, 18 Table 3: outputs 1, 2, 4, 8, 16

Solution: - Table 1: Δ = 3, 3, 3, 3 → linear - Table 2: Δ = 1, 3, 5, 7; Δ² = 2, 2, 2 → quadratic - Table 3: Δ = 1, 2, 4, 8 — not constant; Δ² = 1, 2, 4 — not constant → neither (ratios are constant: it's exponential, wait for Unit 2)

Interpretation: Difference analysis is a two-step decision tree, and "neither" is a legal answer.

Example 2 (medium) — Recover the quadratic 🚫 No-Calc

Problem: f is quadratic with equally spaced values: f(0) = 4, f(1) = 7, f(2) = 14, f(3) = 25. Find f(x) = ax² + bx + c.

Strategy: Second difference gives a; f(0) gives c; one more point gives b.

Solution:

Δ:  3, 7, 11    Δ²: 4, 4  →  2a(1)² = 4  →  a = 2
f(0) = c = 4
f(1) = 2 + b + 4 = 7  →  b = 1
f(x) = 2x² + x + 4

Check with f(3): 2(9) + 3 + 4 = 25 ✓

Interpretation: Constant second differences aren't just a label — they carry the leading coefficient.

Example 3 (medium) — Vertex from zeros, in context 🚫 No-Calc

Problem: A projectile's height is quadratic in time; it launches from the ground at t = 0 and lands at t = 8 seconds, with maximum height 80 m. Find the height function.

Strategy: Zeros at 0 and 8 → factored form h(t) = at(t − 8); vertex at the midpoint of the zeros.

Solution: Vertex at t = 4 by symmetry. h(4) = a(4)(−4) = −16a = 80 → a = −5.

h(t) = −5t(t − 8) = −5t² + 40t

(The Lesson 1 rocket, unmasked.) a = −5 < 0 ✓ concave down, consistent with a maximum.

Interpretation: Zeros → symmetry → vertex is often faster than completing the square.

Example 4 (AP-style) — Equal AROCs, same midpoint 🚫 No-Calc

Problem: g is quadratic, and the average rate of change of g over [1, 5] is 12. What is the average rate of change of g over [2, 4]?

Solution: Both intervals have midpoint 3. For a quadratic, AROC depends only on the interval's midpoint, so the AROC over [2, 4] is also 12.

Interpretation: No formula for g needed. If a problem seems under-determined, a structural property is usually the intended key. (Why it works: AROC of ax²+bx+c over [p, q] is a(p+q) + b, which depends only on p+q — i.e., only on the midpoint.)

Common Mistakes

1. Running differences on unequally spaced inputs. If x jumps 0, 1, 3, 4, raw output differences mean nothing. Fix: check spacing first; if unequal, compute AROCs (Δy/Δx) instead of bare differences.
2. Slope = first difference regardless of step. With x-step 2, first differences are 2m, not m. Fix: always divide by the input step.
3. "Concave up" = "increasing." An upward parabola is decreasing on the entire left half while concave up. Fix (again, it recurs): direction and bend are separate diagnoses.
4. Assuming a table with constant Δ² over four points must be quadratic. It's consistent with quadratic — the honest AP phrasing — since other functions could agree at sampled points. Model-selection FRQs reward "suggests/is consistent with."
5. Sign slip at the vertex formula. x = −b/(2a) — students drop the negative or divide by a instead of 2a. Fix: check that the vertex x-value sits midway between the zeros when zeros are easy.

Practice Problems

12. (FRQ-style) 🚫 The table gives the depth of water D(t), in cm, in a draining tank at equally spaced times.

(i) Show that the data are consistent with a quadratic model. (ii) Find a quadratic model D(t) = at² + bt + c. (iii) Using your model, when (if ever) does the rate of change of depth equal zero, and what does that suggest about the draining process?

FRQ Practice

Full 6-point FRQs begin next lesson. Problem 12 is a 5-point warm-up in FRQ format.