Mathematics Advanced • Year 11 • Module 1 • Lesson 2
Function Notation & Evaluation
Practise HSC-style writing on function evaluation, algebraic substitution, and the difference quotient.
1. Short-answer questions
1.1 For f(x) = 2x² − 5x + 3, evaluate f(4) and f(−2). Show all working with brackets. 3 marks Band 3
1.2 Given g(x) = (x + 3) / (2x − 4):
(a) Find g(5).
(b) Find the value of x for which g(x) is not defined, justifying your answer. 3 marks Band 3-4
1.3 For f(x) = x² + 2x, simplify the difference quotient [f(x + h) − f(x)] / h as far as possible, and state its value as h → 0 at x = 3. 4 marks Band 4
Stuck on 1.3? Expand (x + h)² first, then collect, then divide by h.2. Extended response
2.1 Let f be a polynomial function.
(a) For f(x) = ax² + bx + c (a ≠ 0), simplify the difference quotient [f(x + h) − f(x)] / h, fully expanded in x and h. Hence write down the value as h → 0.
(b) Show that if f is linear (a = 0), the difference quotient is constant and equals b. Interpret this result geometrically.
(c) Hence prove that the difference quotient of a polynomial f(x) = ax² + bx + c, evaluated at x = a₀ as h → 0, equals 2 a a₀ + b, and state in one sentence what this tells us about how the rate of change of a quadratic depends on the position x = a₀. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 4 marks
• 1 mark — correct expansion of f(x + h) = a(x + h)² + b(x + h) + c = ax² + 2axh + ah² + bx + bh + c.
• 1 mark — correct subtraction: f(x + h) − f(x) = 2axh + ah² + bh.
• 1 mark — correct division and simplification: [f(x + h) − f(x)] / h = 2ax + ah + b.
• 1 mark — h → 0 limit: 2ax + b.
Part (b) — 2 marks
• 1 mark — sets a = 0 and derives the quotient = b.
• 1 mark — geometric interpretation: b is the gradient of the line, constant rate of change.
Part (c) — 2 marks
• 1 mark — substitutes x = a₀ into the result of (a), obtains 2 a a₀ + b.
• 1 mark — interpretation: rate of change of a quadratic is itself linear in position (varies with x), unlike the constant rate of a linear function.
Your response:
Stuck? Treat a, b, c as constants. The only variable in the expansion is x and the step is h.How did this worksheet feel?
What I'll revisit before next class:
1.1 — f(4) and f(−2) for f(x) = 2x² − 5x + 3 (3 marks)
Sample response.
f(4) = 2(4)² − 5(4) + 3 = 2(16) − 20 + 3 = 32 − 20 + 3 = 15.
f(−2) = 2(−2)² − 5(−2) + 3 = 2(4) + 10 + 3 = 8 + 10 + 3 = 21.
Marking notes. 1 mark — clear bracket substitution at both inputs. 1 mark — correct f(4) = 15. 1 mark — correct f(−2) = 21. Lose 1 mark for dropping a bracket and ending with (−2)² = −4.
1.2 — g(x) = (x + 3) / (2x − 4) (3 marks)
Sample response.
(a) g(5) = (5 + 3) / (2(5) − 4) = 8 / 6 = 4/3.
(b) g(x) is not defined when the denominator equals 0. 2x − 4 = 0 ⇒ x = 2. At x = 2 we would be dividing by zero, which is undefined; this is therefore excluded from the natural domain.
Marking notes. 1 mark — correct numerical evaluation g(5) = 4/3 (or 1.33...). 1 mark — equation 2x − 4 = 0 solved correctly. 1 mark — explicit justification "division by zero" / "denominator cannot be zero".
1.3 — Difference quotient for f(x) = x² + 2x (4 marks)
Sample response. f(x + h) = (x + h)² + 2(x + h) = x² + 2xh + h² + 2x + 2h.
f(x + h) − f(x) = (x² + 2xh + h² + 2x + 2h) − (x² + 2x) = 2xh + h² + 2h.
[f(x + h) − f(x)] / h = (2xh + h² + 2h) / h = 2x + h + 2 (for h ≠ 0).
As h → 0: limit = 2x + 2. At x = 3: 2(3) + 2 = 8.
Marking notes. 1 mark — correct expansion of (x + h)². 1 mark — correct subtraction with all three remaining terms. 1 mark — correctly simplified quotient 2x + h + 2. 1 mark — correct h → 0 value at x = 3 = 8. Common error: writing (x + h)² = x² + h² (missing 2xh) — loses 2/4.
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). Take f(x) = ax² + bx + c with a ≠ 0. Then
f(x + h) = a(x + h)² + b(x + h) + c = a(x² + 2xh + h²) + bx + bh + c = ax² + 2axh + ah² + bx + bh + c. [1 mark — full expansion of f(x + h).]
Subtracting f(x):
f(x + h) − f(x) = (ax² + 2axh + ah² + bx + bh + c) − (ax² + bx + c) = 2axh + ah² + bh = h(2ax + ah + b). [1 mark — correct subtraction.]
Dividing by h (h ≠ 0):
[f(x + h) − f(x)] / h = 2ax + ah + b. [1 mark — clean quotient.]
As h → 0, the term ah → 0, leaving
lim [f(x + h) − f(x)] / h = 2ax + b. [1 mark — h → 0 limit stated.]
Part (b). Setting a = 0, f(x) = bx + c. From (a) the quotient reduces to 2(0)x + 0 + b = b, constant in both x and h. [1 mark — derivation of constant quotient.]
Geometrically, b is the gradient of the straight line y = bx + c, and the difference quotient is precisely the slope between any two points on the line — which, for a line, is the same everywhere. So a linear function has constant rate of change b. [1 mark — geometric interpretation.]
Part (c). Substituting x = a₀ into the result of (a):
[f(a₀ + h) − f(a₀)] / h → 2 a a₀ + b as h → 0. [1 mark — substitution.]
Unlike the linear case in (b), the limit depends on a₀: the instantaneous rate of change of a quadratic is itself a linear function of position. So a parabola's slope grows (or shrinks) as we move along the x-axis, whereas a straight line's slope is the same everywhere. [1 mark — qualitative interpretation.] ▮
Total: 8/8.
Band descriptors for marker.
Band 3: Attempts the expansion but drops the 2xh term, or fails to subtract cleanly. Doesn't reach the h → 0 step. ≈ 2-3 marks.
Band 4: Completes (a) up to the divided form 2ax + ah + b but doesn't push to h → 0; (b) shows the constant b but no interpretation; misses (c). ≈ 4-5 marks.
Band 5: Completes (a) including h → 0 = 2ax + b. (b) and (c) attempted but interpretation is loose or omitted in one part. ≈ 6-7 marks.
Band 6: All three parts complete, with explicit limit notation, geometric interpretation of b in (b), and a clear sentence in (c) contrasting quadratic vs linear rates of change. 8/8.