Mathematics Advanced • Year 11 • Module 4 • Lesson 6

Graphs of Logarithmic Functions

Build procedural fluency in identifying key features of y = log_ax and its translations: asymptote, x-intercept, domain.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the standard features of the basic graph y = log_ax (with a > 1):

Vertical asymptote: x = ____________ x-intercept: ( ____ , ____ )

Domain: ____________ Range: ____________

Q1.2 The graph y = log_ax is the reflection of y = a^x across the line y = ____________.

Q1.3 For y = log_a(x − h) + k, state the asymptote and the point that replaces (1, 0):

Asymptote: x = ____________ Point: ( ____________ , ____________ )

Stuck? Revisit lesson § Concept and § Formula Reference.

2. Worked example — sketching y = log₂(x − 1) + 1

Each step asks "what does this transformation do?" Read the right-hand reason after each line.

Problem. Sketch y = log₂(x − 1) + 1, stating the asymptote, x-intercept, and one extra point.

Step 1 — Identify base and transformations.

Base function: y = log₂x Shifts: right by 1, up by 1

Reason: (x − 1) inside the log is a horizontal shift right by 1; the "+ 1" outside is a vertical shift up.

Step 2 — Move the vertical asymptote.

x − 1 = 0 ⇒ x = 1 (was x = 0)

Reason: the asymptote of y = log₂x is where the argument equals 0.

Step 3 — Find the x-intercept (set y = 0).

0 = log₂(x − 1) + 1 ⇒ log₂(x − 1) = −1

x − 1 = 2⁻¹ = ½ ⇒ x = 1.5 ∴ (1.5, 0)

Reason: convert from log form to exponential form.

Step 4 — Find a clean extra point.

At x = 2: y = log₂(1) + 1 = 0 + 1 = 1 ∴ (2, 1)

Reason: choose x − 1 = 1 so the log gives 0.

Conclusion. Asymptote x = 1; passes through (1.5, 0) and (2, 1); slow-rising shape approaching x = 1 from the right.

3. Faded example — fill in the missing steps

Sketch y = log₃(x + 2) by completing each blank. 4 marks

Step 1 — Asymptote (argument = 0):

x + 2 = 0 ⇒ x = ____________

Step 2 — Domain (argument > 0):

x + 2 > 0 ⇒ x > ____________

Step 3 — x-intercept (set y = 0):

log₃(x + 2) = 0 ⇒ x + 2 = ____________ ⇒ x = ____________

Step 4 — Clean extra point (make the argument = base):

x + 2 = 3 ⇒ x = ____________; y = log₃(3) = ____________. Point: ( ____ , ____ )

Conclusion. Sketch an increasing curve approaching x = ____________ from the right, through ( ____ , ____ ) and ( ____ , ____ ).

Stuck? Revisit lesson § Worked example 2 — y = log₂(x − 1) + 1.

4. Graduated practice — state asymptote, x-intercept, domain

For each function, give the vertical asymptote, the x-intercept (as an ordered pair), and the domain.

Foundation — basic and reflected logs (4 questions)

Q	Function	Asymptote	x-intercept	Domain
4.1 1	y = log₂x
4.2 1	y = ln x
4.3 1	y = log₅x
4.4 1	y = −log₃x

Standard — horizontal and vertical shifts (6 questions)

Show the line of working: argument = 0 for the asymptote; y = 0 for the x-intercept.

4.5 y = log₂(x − 3) 2 marks

4.6 y = ln(x + 4) 2 marks

4.7 y = log₂(x + 1) − 1 2 marks

4.8 y = log₃(x − 2) + 2 2 marks

4.9 y = ln(x − 1) + 2 2 marks

4.10 y = 2 log₂(x) − 4 2 marks

Extension — recover the equation from features (2 questions)

4.11 A logarithmic curve has vertical asymptote x = 2 and passes through (3, 0) and (5, 1). Find its equation in the form y = log_a(x − h). 3 marks

4.12 The curve y = ln(x + a) + b passes through (0, 0) and (e − 1, 1). Find a and b. 3 marks

Stuck on 4.11? Substitute each point into y = log_a(x − 2) and solve for a.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 (y = log₂x) I have asymptote x = 0, x-intercept (1, 0), domain x > 0 — all three are needed.

For 4.2 (y = ln x) I noted ln is shorthand for log_e, so the features are identical to 4.1.

For 4.4 (y = −log₃x) the negative reflects the curve in the x-axis but does not change the asymptote or x-intercept.

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Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Standard features of y = log_ax

Asymptote: x = 0. x-intercept: (1, 0). Domain: x > 0. Range: all real numbers.

Q1.2 — Reflection line

y = log_ax is the reflection of y = a^x across y = x. Logarithmic and exponential functions are inverses.

Q1.3 — Transformed log features

For y = log_a(x − h) + k: asymptote x = h; the point (1, 0) on the base graph maps to (1 + h, k) (set argument = 1 so the log = 0, then add k).

Q3 — Faded example y = log₃(x + 2)

Step 1: x = −2. Step 2: x > −2. Step 3: x + 2 = 1, x = −1. Step 4: x = 1, y = 1, point (1, 1).
Conclusion: increasing curve approaching x = −2 from the right, through (−1, 0) and (1, 1).

Q4.1 — y = log₂x

Asymptote x = 0. x-intercept (1, 0). Domain x > 0. The base case — every other graph in this section is a transformation of this one.

Q4.2 — y = ln x

ln = log_e. Asymptote x = 0. x-intercept (1, 0). Domain x > 0. Identical features to 4.1.

Q4.3 — y = log₅x

Asymptote x = 0. x-intercept (1, 0). Domain x > 0. Changing the base only affects how fast the curve rises, not the asymptote, x-intercept or domain.

Q4.4 — y = −log₃x

Asymptote x = 0. x-intercept (1, 0) (because −log₃(1) = 0). Domain x > 0. The minus reflects the curve in the x-axis so the graph is now decreasing, but the asymptote/intercept/domain are unchanged.

Q4.5 — y = log₂(x − 3)

Argument = 0: x − 3 = 0, asymptote x = 3. Set y = 0: log₂(x − 3) = 0 ⇒ x − 3 = 1, so x-intercept (4, 0). Domain x > 3.

Q4.6 — y = ln(x + 4)

Asymptote x = −4. Set y = 0: x + 4 = 1, x-intercept (−3, 0). Domain x > −4.

Q4.7 — y = log₂(x + 1) − 1

Asymptote x = −1. Set y = 0: log₂(x + 1) = 1 ⇒ x + 1 = 2, x-intercept (1, 0). Domain x > −1.

Q4.8 — y = log₃(x − 2) + 2

Asymptote x = 2. Set y = 0: log₃(x − 2) = −2 ⇒ x − 2 = 3⁻² = 1/9, x-intercept (2 + 1/9, 0) = (19/9, 0). Domain x > 2.

Q4.9 — y = ln(x − 1) + 2

Asymptote x = 1. Set y = 0: ln(x − 1) = −2 ⇒ x − 1 = e⁻², x-intercept (1 + e⁻², 0) ≈ (1.135, 0). Domain x > 1.

Q4.10 — y = 2 log₂(x) − 4

Asymptote x = 0 (a vertical stretch by factor 2 does not move the asymptote). Set y = 0: 2 log₂(x) = 4 ⇒ log₂(x) = 2 ⇒ x = 4, x-intercept (4, 0). Domain x > 0.

Q4.11 — Recover from features (asymptote x = 2, points (3, 0) and (5, 1))

Asymptote x = 2 ⇒ form y = log_a(x − 2) + k.
Substitute (3, 0): 0 = log_a(1) + k = 0 + k, so k = 0.
Substitute (5, 1): 1 = log_a(3), so a¹ = 3, giving a = 3.
Equation: y = log₃(x − 2).

Q4.12 — y = ln(x + a) + b through (0, 0) and (e − 1, 1)

From (0, 0): 0 = ln(a) + b. …(i)
From (e − 1, 1): 1 = ln(e − 1 + a) + b. …(ii)
Subtract (i) from (ii): 1 = ln(e − 1 + a) − ln(a) = ln((e − 1 + a)/a).
So (e − 1 + a)/a = e ⇒ e − 1 + a = ea ⇒ e − 1 = ea − a = a(e − 1) ⇒ a = 1.
Then from (i): b = −ln(1) = 0. Equation: y = ln(x + 1).