Mathematics Advanced • Year 11 • Module 4 • Lesson 2
Graphs of Exponential Functions
Build fluency in sketching y = ax, identifying asymptotes and intercepts, and applying transformations.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 The graph of y = ax always passes through the point ( ____ , ____ ), because a0 = ____.
Q1.2 Fill in the asymptote rule for vertical translations:
The graph of y = ax + k has horizontal asymptote y = ____ . The graph of y = ax − h has horizontal asymptote y = ____.
Q1.3 Match each transformation to its effect on y = 2x:
A. y = −2x B. y = 2−x C. y = 2x + 3 D. y = 2x − 1
(i) shift left 1 / shift right 1 (circle one) (ii) shift up 3 (iii) reflect in x-axis (iv) reflect in y-axis
2. Worked example — sketch y = 2x showing key features
Follow each line.
Problem. Sketch y = 2x, labelling the asymptote, the y-intercept, and one other point.
Step 1 — Identify the asymptote.
As x → −∞, 2x → 0. ∴ asymptote: y = 0 (x-axis).
Reason: the basic exponential has the x-axis as its asymptote.
Step 2 — Find the y-intercept.
x = 0: y = 20 = 1. ∴ (0, 1).
Step 3 — Plot one more guide point.
x = 1: y = 2. ∴ (1, 2). (Also (−1, 1/2) for shape.)
Step 4 — Draw a smooth increasing curve.
Approaches y = 0 from above as x → −∞; rises steeply for x > 1. Dashed asymptote labelled y = 0.
3. Faded example — sketch y = 3x − 1 + 2
Fill in the missing pieces. 4 marks
Step 1 — Start with y = 3x.
Transformations applied: shift ____________ by ____ unit(s), then shift ____________ by ____ unit(s).
Step 2 — New asymptote.
The asymptote of y = 3x is y = ____ . After the vertical shift, asymptote: y = ____.
Step 3 — y-intercept.
x = 0: y = 30 − 1 + 2 = 3−1 + 2 = ____ + 2 = ____. y-intercept: (0, ____).
Step 4 — Range.
Since the basic range y > 0 is shifted up by 2, the new range is y ____ ____.
4. Graduated practice
State features for each function. Sketch on graph paper where indicated.
Foundation — read features off the equation (4 questions)
| Q | Function | Asymptote | y-intercept |
|---|---|---|---|
| 4.1 1 | y = 2x | ||
| 4.2 1 | y = 3x − 1 | ||
| 4.3 1 | y = (1/2)x | ||
| 4.4 1 | y = 2x + 5 |
Standard — sketch with transformations (6 questions)
For each, state asymptote, y-intercept, and one extra point. Then sketch.
4.5 y = 2x − 2. 2 marks
4.6 y = 3x + 4. 2 marks
4.7 y = −2x. (Reflection in x-axis.) State range. 2 marks
4.8 y = 2−x. (Reflection in y-axis.) Is it growth or decay? 2 marks
4.9 y = (1/2)x − 3. State asymptote, y-intercept and range. 2 marks
4.10 y = 3 · 2x. (Vertical dilation factor 3.) State y-intercept and asymptote. 2 marks
Extension — equations from features (2 questions)
4.11 A curve has equation y = a · bx and passes through (0, 5) and (2, 20). Find a and b, showing both substitutions. 3 marks
4.12 A graph has equation y = ax + k, asymptote y = 2, and passes through (0, 3). Find a if the graph also passes through (1, 4). 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Common point
(0, 1), because a0 = 1.
Q1.2 — Asymptotes
y = ax + k has asymptote y = k. y = ax − h has asymptote y = 0 (a horizontal shift does not move the horizontal asymptote).
Q1.3 — Match transformations
A → (iii) reflect in x-axis. B → (iv) reflect in y-axis. C → (ii) shift up 3. D → shift right 1 (subtraction inside the exponent shifts right).
Q3 — Faded example y = 3x − 1 + 2
Step 1: shift right by 1, then shift up by 2.
Step 2: original asymptote y = 0; new asymptote y = 2.
Step 3: y = 3−1 + 2 = 1/3 + 2 = 7/3. y-intercept: (0, 7/3).
Step 4: range y > 2.
Q4.1–4.4 — Features
4.1: asymptote y = 0, y-int (0, 1). 4.2: asymptote y = −1, y-int (0, 0) (since 30 − 1 = 0). 4.3: asymptote y = 0, y-int (0, 1). 4.4: asymptote y = 5, y-int (0, 6).
Q4.5 — y = 2x − 2
Shift right 2. Asymptote y = 0. y-int (0, 2−2) = (0, 1/4). Extra point: (2, 1). Increasing curve.
Q4.6 — y = 3x + 4
Shift up 4. Asymptote y = 4. y-int (0, 30 + 4) = (0, 5). Extra point (1, 7).
Q4.7 — y = −2x
Reflection of y = 2x in the x-axis. Asymptote y = 0. y-int (0, −1). Range y < 0. Curve sits entirely below the x-axis.
Q4.8 — y = 2−x
Equivalent to y = (1/2)x, which is decay. Asymptote y = 0; y-int (0, 1). Decreasing curve.
Q4.9 — y = (1/2)x − 3
Asymptote y = −3. y-int (0, 1 − 3) = (0, −2). Range y > −3. Decreasing curve approaches y = −3 from above as x → ∞.
Q4.10 — y = 3 · 2x
Vertical dilation by factor 3. Asymptote y = 0 (unchanged). y-int (0, 3) (since 3 · 20 = 3). Extra point (1, 6).
Q4.11 — Find a and b in y = a · bx
At (0, 5): 5 = a · b0 = a, so a = 5. At (2, 20): 20 = 5 · b2, so b2 = 4, giving b = 2 (positive root since base is positive).
Q4.12 — Find a in y = ax + k
Asymptote y = 2 → k = 2. So y = ax + 2. At (0, 3): 3 = a0 + 2 = 1 + 2 = 3 ✓ (gives no new information — this point is automatically on every shift y = ax + 2). At (1, 4): 4 = a + 2 ⇒ a = 2. Equation: y = 2x + 2.