Mathematics Advanced • Year 11 • Module 4 • Lesson 1
Introduction to Exponential Functions
Practise HSC-style writing on the definition, properties and modelling use of exponential functions.
1. Short-answer questions
1.1 Evaluate f(x) = (1/2)x at x = −3, 0, 4. Show one line of working for each. 3 marks Band 3
1.2 The value of a piece of office equipment is modelled by V(t) = 8000 × (0.8)t, where t is the number of years from purchase. Find V(3) to the nearest dollar, and state whether the model represents growth or decay (justify with a single phrase about the base). 3 marks Band 3-4
1.3 For f(x) = ax with a > 1:
(a) State the range of f.
(b) Explain in one sentence why f(x) > 0 for every real x.
(c) Describe the behaviour of f(x) as x → −∞. 4 marks Band 4
2. Extended response
2.1 A biology student measures a yeast colony at t = 0 hours and finds 200 cells. Two hours later there are 800 cells. They believe the colony obeys an exponential model of the form P(t) = A × bt, where t is in hours.
(a) Find the values of A and b, showing your working.
(b) Use your model to predict the population at t = 5 hours, to the nearest whole cell.
(c) Show algebraically that, under this model, the population doubles every hour.
(d) The student observes that at t = 10 hours the actual count is only 95 000 cells, not what the model predicts. Identify one realistic biological reason why an unconstrained exponential growth model eventually becomes a poor predictor for a real colony. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — identifies A = P(0) = 200 from the initial condition.
• 1 mark — substitutes (t, P) = (2, 800) into 200·b2 = 800, solves b2 = 4, concludes b = 2 (rejecting b = −2 because base must be positive).
Part (b) — 1 mark
• 1 mark — P(5) = 200 × 25 = 200 × 32 = 6 400 cells.
Part (c) — 2 marks
• 1 mark — writes P(t + 1) = 200 × 2t+1.
• 1 mark — shows P(t + 1) = 200 × 2t × 2 = 2 × P(t), with a concluding sentence "population doubles each hour".
Part (d) — 2 marks
• 1 mark — identifies a specific limiting factor (e.g. nutrient depletion, build-up of waste, restricted physical volume, immune/predator response). 1 mark — connects this to the model: an unconstrained exponential predicts unbounded growth, but real populations are bounded by environment, so the model loses fit at large t.
Your response:
Stuck on (c)? Use the index law 2t+1 = 2t · 21.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Evaluate (1/2)x (3 marks)
Sample response.
f(−3) = (1/2)−3 = 23 = 8.
f(0) = (1/2)0 = 1.
f(4) = (1/2)4 = 1/24 = 1/16.
Marking notes. 1 mark per value with one line of working. The most common mark loss is at f(−3): students forget that the negative exponent flips the fraction, so (1/2)−3 = 23 = 8 (not −1/8 or −8).
1.2 — V(t) = 8000 × (0.8)t (3 marks)
Sample response. V(3) = 8000 × (0.8)3 = 8000 × 0.512 = $4 096. The base 0.8 satisfies 0 < 0.8 < 1, so the model represents exponential decay.
Marking notes. 1 mark for (0.8)3 = 0.512. 1 mark for the correct dollar value. 1 mark for "decay because 0 < base < 1" with the base value cited. A bare "decay" with no reference to the base scores only 0.5.
1.3 — Properties of f(x) = ax, a > 1 (4 marks)
Sample response.
(a) Range: y > 0 (or (0, ∞)).
(b) For any a > 0 and any real x, ax is positive — multiplying a positive base by itself any number of times (or taking its reciprocal) never produces a negative or zero value.
(c) As x → −∞, write ax = 1/a−x. Then −x → +∞, and since a > 1 we have a−x → ∞, so 1/a−x → 0. Hence f(x) → 0+ (the x-axis is a horizontal asymptote).
Marking notes. (a) 1 mark for stating range strictly > 0. (b) 1 mark for explaining positivity via "positive base × positive base is positive" or the reciprocal argument. (c) 2 marks: 1 for rewriting ax in a form that makes the limit obvious (e.g. 1/a−x), 1 for concluding the limit is 0. A purely numerical answer like "very small" without an algebraic rewrite earns only 1 of the 2 limit marks.
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). The model is P(t) = A · bt. Applying the initial condition P(0) = 200:
A · b0 = A · 1 = A = 200. [1 mark — A identified.]
Applying P(2) = 800:
200 · b2 = 800 ⇒ b2 = 4 ⇒ b = 2 (rejecting b = −2 because the base of an exponential must be positive). [1 mark — b correctly identified, with rejection of negative root.]
Therefore the model is P(t) = 200 · 2t.
Part (b). P(5) = 200 · 25 = 200 · 32 = 6 400 cells. [1 mark.]
Part (c). Using the index law:
P(t + 1) = 200 · 2t + 1 = 200 · 2t · 2 = 2 · (200 · 2t) = 2 · P(t). [1 mark — sets up P(t+1); 1 mark — concludes "population doubles each hour" with the algebraic ratio shown.]
Part (d). A real yeast colony is bounded by environmental constraints — for example, nutrient depletion: in a closed flask the sugar supply runs out, so once roughly half the nutrients are consumed, the per-cell growth rate slows and eventually halts. The exponential model P(t) = 200 · 2t implicitly assumes nutrients (and space) are unlimited, so it correctly fits the early "log-phase" growth but overpredicts as t grows. By t = 10 hours the model predicts 200 · 1024 = 204 800 cells; the measured 95 000 reflects the colony entering the resource-limited "stationary phase". [1 mark — names a specific limiting factor; 1 mark — links to model assumption and concludes the model overpredicts at large t.]
Total: 7/7.
Band descriptors for marker.
Band 3: Identifies A = 200; sets up but does not solve for b; some attempt at (b). Often skips (c) and (d). ≈ 2 marks.
Band 4: Correct A and b; correct P(5). Attempts (c) but writes "the model doubles" without algebra. (d) gives a vague "growth slows" answer with no specific cause. ≈ 4 marks.
Band 5: Full (a)(b)(c). (d) names a limiting factor but does not connect it to the model assumption. ≈ 5-6 marks.
Band 6: All parts complete. (a) rejects negative root with reason. (c) uses index laws cleanly to show P(t+1) = 2P(t). (d) names a specific biological cause and identifies the unrealistic assumption (unbounded resources) that breaks the model. 7/7.