Mathematics Advanced • Year 11 • Module 4 • Lesson 2

Graphs of Exponential Functions

Practise HSC-style writing on sketching, transformations and finding equations from features.

Master · Past-Paper Style

1. Short-answer questions

1.1 Sketch y = 2^{x − 1} + 1, labelling the asymptote, the y-intercept, and one extra point. 3 marks Band 3

1.2 An exponential curve has equation y = a · b^x and passes through the points (0, 4) and (3, 32). Find a and b, showing your working. 3 marks Band 3-4

1.3 The curve y = a^x + k has horizontal asymptote y = −1 and passes through the point (0, 0).
(a) Find k.
(b) Explain why the point (0, 0) does not by itself determine a uniquely, given k. What additional information would you need? 4 marks Band 4

Stuck on 1.3(b)? Show that (0, 0) follows automatically once k = −1.

2. Extended response

2.1 A research lab models the rebound height of a bouncing ball as an exponential decay: after each bounce, the maximum height of the next bounce is a fixed fraction of the previous height. The ball is dropped from h₀ = 200 cm and the second bounce reaches 128 cm.
(a) Show that the model H(n) = h₀ · rⁿ, where n is the bounce number, gives r² = 0.64, and find r.
(b) State the maximum height H(1) after the first bounce and the height H(5) after the fifth bounce. Round H(5) to the nearest cm.
(c) On the same axes, sketch y = H(n) and the horizontal line H = 10 cm (the "stop" threshold). Identify the smallest whole number of bounces n for which H(n) < 10.
(d) Describe one specific transformation of the basic graph y = r^x (with r = 0.8) that produces the graph of y = H(n) above, and state how the asymptote behaves in both. 7 marks Band 5-6

Explicit marking criteria

Part (a) — 2 marks

• 1 mark — substitutes H(2) = 128, h₀ = 200 into the model and obtains r² = 0.64.

• 1 mark — solves r = 0.8 (positive root justified: heights are positive, so the multiplier per bounce is positive).

Part (b) — 2 marks

• 1 mark — H(1) = 200 · 0.8 = 160 cm.

• 1 mark — H(5) = 200 · 0.8⁵ = 200 · 0.32768 ≈ 66 cm.

Part (c) — 2 marks

• 1 mark — sketch shows decreasing exponential through (0, 200), (1, 160), (5, 66) with H = 10 line drawn.

• 1 mark — by trial: H(13) = 200 · 0.8¹³ ≈ 11.5 cm; H(14) ≈ 9.2 cm. First n with H(n) < 10 is n = 14.

Part (d) — 1 mark

• 1 mark — H(n) = 200 · 0.8ⁿ is a vertical dilation of y = 0.8ⁿ by factor 200 (the y-coordinate of each point is multiplied by 200). The horizontal asymptote remains y = 0 in both, because vertical dilation does not move the asymptote.

Your response:

Stuck on uniqueness in 1.3(b) earlier? Apply the same idea: a single point doesn't always pin a parameter once a constraint is satisfied automatically.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — Sketch y = 2^{x − 1} + 1 (3 marks)

Sample response.
Asymptote: y = 1 (from the +1 outside the exponent).
y-intercept at x = 0: y = 2⁻¹ + 1 = 1/2 + 1 = 3/2, so (0, 3/2).
Extra point at x = 1: y = 2⁰ + 1 = 2, so (1, 2).
Sketch: increasing curve, asymptote labelled (dashed) at y = 1, passing through (0, 3/2) and (1, 2), rising steeply for x > 1.

Marking notes. 1 mark — asymptote labelled at y = 1 (not y = 0). 1 mark — correct y-intercept (0, 3/2). 1 mark — correct overall shape (increasing, approaching y = 1 from above as x → −∞). A sketch that draws the curve crossing the asymptote scores at most 2/3.

1.2 — Find a and b in y = a · b^x through (0, 4), (3, 32) (3 marks)

Sample response.
At (0, 4): 4 = a · b⁰ = a, so a = 4.
At (3, 32): 32 = 4 · b³, so b³ = 8, giving b = 2.
Equation: y = 4 · 2^x.

Marking notes. 1 mark — uses the y-intercept (0, 4) correctly to obtain a = 4. 1 mark — substitutes (3, 32) and reaches b³ = 8. 1 mark — concludes b = 2 (positive cube root). Students who try to take logs unnecessarily score full marks if correct, but the "match y-intercept first" approach is cleanest.

1.3 — y = a^x + k, asymptote y = −1, point (0, 0) (4 marks)

Sample response.
(a) The asymptote of y = a^x + k is y = k, so k = −1.
(b) With k = −1, the equation becomes y = a^x − 1. At x = 0 this gives y = a⁰ − 1 = 1 − 1 = 0, automatically — so (0, 0) is on the graph for every valid base a. The point gives no information beyond what k already provides. To pin a uniquely we would need a second non-zero data point, e.g. the value at x = 1 or x = 2.

Marking notes. (a) 1 mark — k = −1 from the asymptote rule. (b) 1 mark — substitutes x = 0 and shows (0, 0) is forced once k = −1. 1 mark — explicit statement that a is therefore undetermined by the given data. 1 mark — proposes a specific second piece of information (a coordinate at any other x) to pin a. Common error: students set 0 = a + k and "solve" for a in terms of k; this is consistent but does not address the multiplicity.

2.1 — Extended response (7 marks): sample Band-6 response with annotations

Sample Band-6 response.

Part (a). Model H(n) = 200 · rⁿ. At n = 2, H(2) = 128:

200 · r² = 128 ⇒ r² = 128/200 = 0.64. [1 mark — r² = 0.64.]

Taking the positive square root (heights are positive, so r > 0):

r = √0.64 = 0.8. [1 mark — r = 0.8 with reason.]

Part (b). H(1) = 200 · 0.8¹ = 160 cm. [1 mark.]

H(5) = 200 · 0.8⁵ = 200 · 0.32768 = 65.536 ≈ 66 cm. [1 mark.]

Part (c). Sketch: decreasing exponential decay, asymptote y = 0 (dashed), passing through (0, 200), (1, 160), (5, 66); a horizontal line H = 10 drawn across the plot. [1 mark — sketch with both curves and the threshold line.]

By trial: H(13) = 200 · 0.8¹³ = 200 · 0.05498 ≈ 11.0 cm (≥ 10); H(14) = 200 · 0.8¹⁴ ≈ 8.8 cm (< 10). So the first bounce number with H(n) < 10 is n = 14. [1 mark.]

Part (d). H(n) = 200 · 0.8ⁿ is the basic graph y = 0.8ⁿ dilated vertically by factor 200 (each y-value is multiplied by 200). The horizontal asymptote of both graphs is y = 0, because a vertical dilation does not move the horizontal asymptote (only vertical translations do). [1 mark.]

Total: 7/7.

Band descriptors for marker.

Band 3: Identifies model and computes H(1). Does not solve cleanly for r (e.g. writes 200r = 128). ≈ 2-3 marks.

Band 4: Correct r and both H(1), H(5). Sketch has correct shape but no threshold line. Skips or guesses at (c) trial. (d) gives vague "shrinks" answer. ≈ 4-5 marks.

Band 5: All parts attempted with correct numerical answers. (d) identifies vertical dilation but does not comment on the asymptote. ≈ 5-6 marks.

Band 6: Full proof in (a) with positive-root justification. Sketch labels asymptote and both curves. (c) trial shown explicitly. (d) names the specific transformation and correctly states the asymptote remains y = 0 with reasoning. 7/7.