Mathematics Advanced • Year 11 • Module 4 • Lesson 6
Graphs of Logarithmic Functions
Past-paper-style problems on sketching, recovering an equation from features, and writing a structured response on inverse-graph reflection.
1. Short-answer questions
1.1 Sketch y = log2(x + 1) − 1. On your sketch, label the vertical asymptote, the x-intercept, and one additional clean point with integer coordinates. 3 marks Band 3-4
1.2 A logarithmic function f has vertical asymptote x = 1 and passes through the points (2, 0) and (5, 2). Find the equation of f in the form f(x) = loga(x − h) + k, showing all working. 3 marks Band 4
1.3 State the domain and range of y = ln(3 − x) and find its x-intercept. Justify the domain from the requirement that the argument of the log must be positive. 4 marks Band 4
Stuck on 1.3? "3 − x > 0" gives the domain; the asymptote is at 3 − x = 0.2. Extended response
2.1 Consider the curves C1: y = ex − 1 and C2: y = ln(x) + 1.
(a) Show that C2 is the inverse function of C1 by finding the inverse of y = ex − 1 algebraically.
(b) State the line of reflection that maps C1 onto C2, and verify that the point (1, 1) is a fixed point of this reflection.
(c) State the asymptote of each curve, and explain how the horizontal asymptote of C1 and the vertical asymptote of C2 correspond under the reflection.
(d) Sketch both curves on the same axes, showing the line of reflection. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — swaps x and y, then isolates the exponential: x = ey − 1.
• 1 mark — takes ln of both sides and solves: y = ln(x) + 1.
Part (b) — 2 marks
• 1 mark — names the line y = x.
• 1 mark — verifies (1, 1) lies on both curves and on y = x.
Part (c) — 2 marks
• 1 mark — states both asymptotes: C1: y = 0; C2: x = 0.
• 1 mark — explains that reflection in y = x swaps the roles of x and y, sending the horizontal asymptote y = 0 of C1 to the vertical asymptote x = 0 of C2.
Part (d) — 1 mark
• 1 mark — sketch shows: C1 above y = 0 with horizontal asymptote, C2 to the right of x = 0 with vertical asymptote, the line y = x dashed, and both curves passing through (1, 1).
Your response:
Stuck on (c)? Reflection in y = x swaps (a, b) ↔ (b, a), so horizontal lines (y = constant) become vertical lines (x = constant).How did this worksheet feel?
What I'll revisit before next class:
1.1 — Sketch y = log2(x + 1) − 1 (3 marks)
Sample response. Asymptote: x + 1 = 0 ⇒ x = −1. x-intercept: log2(x + 1) = 1 ⇒ x + 1 = 2 ⇒ (1, 0). Extra point at x = 0: y = log2(1) − 1 = 0 − 1 = −1, so (0, −1). Sketch: increasing curve approaching x = −1 from the right, through (0, −1) and (1, 0), rising slowly thereafter.
Marking notes. 1 mark — asymptote x = −1 (labelled). 1 mark — x-intercept (1, 0) (labelled). 1 mark — correct increasing shape with one extra labelled point. Missing asymptote line or unlabelled features = deduct 0.5 each.
1.2 — Recover equation from (2, 0) and (5, 2), asymptote x = 1 (3 marks)
Sample response. Asymptote x = 1 ⇒ form f(x) = loga(x − 1) + k. Substitute (2, 0): 0 = loga(1) + k = 0 + k, so k = 0. Substitute (5, 2): 2 = loga(4), so a2 = 4, giving a = 2 (rejecting a = −2 since the base must be positive). f(x) = log2(x − 1).
Marking notes. 1 mark — correct form with horizontal shift. 1 mark — k = 0 from (2, 0). 1 mark — a = 2 from (5, 2). A student who writes "a = ±2" without rejecting the negative base scores 0.5 on the last mark.
1.3 — y = ln(3 − x): domain, range, x-intercept (4 marks)
Sample response. Domain: argument > 0 ⇒ 3 − x > 0 ⇒ x < 3, i.e. (−∞, 3). Range: as x → 3−, 3 − x → 0+, so ln(3 − x) → −∞; as x → −∞, 3 − x → +∞, so ln(3 − x) → +∞. Range: all real numbers. x-intercept: ln(3 − x) = 0 ⇒ 3 − x = 1 ⇒ x = 2, so (2, 0).
Marking notes. 1 mark — domain x < 3 with the justification "3 − x > 0". 1 mark — range all real numbers. 1 mark — x-intercept (2, 0). 1 mark — correct working shown for the x-intercept (sets y = 0, exponentiates). Common error: students give domain x > 3 because they default to "argument is a shift of x".
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). Start with y = ex − 1. Swap x and y to find the inverse:
x = ey − 1. [1 mark — swap performed.]
Take the natural logarithm of both sides:
ln(x) = y − 1 ⇒ y = ln(x) + 1. [1 mark — solved for y.]
This is exactly C2, so C2 is the inverse function of C1.
Part (b). The reflection that maps a function onto its inverse is reflection in the line y = x. [1 mark.]
Verify (1, 1) is fixed: on C1, e1 − 1 = e0 = 1 ✓. On C2, ln(1) + 1 = 0 + 1 = 1 ✓. Both curves pass through (1, 1), and (1, 1) lies on y = x (since y-coordinate equals x-coordinate), so the reflection in y = x leaves (1, 1) fixed. [1 mark — verified on both curves AND on y = x.]
Part (c). C1: y = ex − 1 has horizontal asymptote y = 0 (since ex − 1 → 0 as x → −∞). C2: y = ln(x) + 1 has vertical asymptote x = 0 (since ln(x) → −∞ as x → 0+). [1 mark — both stated.]
Reflection in y = x swaps the roles of x and y: a horizontal line y = c maps to the vertical line x = c. The horizontal asymptote y = 0 of C1 therefore maps to the vertical asymptote x = 0 of C2. The geometric picture: the curve approaching y = 0 from above (as x → −∞) reflects to a curve approaching x = 0 from the right (as y → −∞), which is precisely the behaviour of ln(x) + 1 near x = 0. [1 mark — explanation invokes "swap x and y".]
Part (d). Sketch: dashed line y = x diagonally; C1 (the exponential) lying entirely above y = 0, passing through (0, e−1) and (1, 1), with horizontal asymptote y = 0; C2 (the log) lying entirely to the right of x = 0, passing through (e−1, 0) and (1, 1), with vertical asymptote x = 0; the two curves are mirror images across y = x. [1 mark.]
Total: 7/7.
Band descriptors for marker.
Band 3: Finds the inverse algebraically but does not link it to "reflection in y = x"; may name the line of reflection without verifying (1, 1); states asymptotes without explaining the correspondence. ≈ 3-4 marks.
Band 4: Inverse found correctly, line of reflection named, asymptotes stated; verifies (1, 1) on at most one curve; sketch present but missing the line y = x or one asymptote. ≈ 4-5 marks.
Band 5: All parts complete; sketch labels all key features (asymptotes, intercepts, line y = x); explanation in (c) mentions "swap x and y" but lacks the closing geometric picture. ≈ 5-6 marks.
Band 6: Full working in (a); reflection line named with verified fixed point (1, 1) and noted to lie on y = x; both asymptotes correctly identified with an explicit explanation of the swap; sketch labelled with all features; uses precise language ("inverse function", "reflection in y = x", "fixed point"). 7/7.