Mathematics Advanced • Year 11 • Module 1 • Lesson 13
Sketching & Modelling Transformed Functions
Build the key-feature method: track vertex, intercepts and asymptotes to sketch transformed graphs without plotting a table of values.
1. Quick recall
Answer each part in the space provided. 1 mark each
Q1.1 List the five key features you should always identify before sketching a transformed function:
1. _______________________ 2. _______________________ 3. _______________________
4. _______________________ 5. _______________________
Q1.2 A turning point (x_0, y_0) on y = f(x) becomes (___________ , ___________) on y = a f(b(x − h)) + k.
Q1.3 A vertical asymptote x = c on y = f(x) becomes x = _______________ on y = a f(b(x − h)) + k.
A horizontal asymptote y = d becomes y = _______________.
2. Worked example — sketching y = −2(x − 3)² + 4
Follow each step. The method tracks the parent's anchor points instead of plotting a table.
Problem. Sketch the graph of y = −2(x − 3)² + 4, labelling the vertex and x-intercepts.
Step 1 — Identify the parent function.
Parent: y = x² (parabola, vertex (0, 0), opens upward).
Step 2 — Read the transformations.
a = −2, b = 1, h = 3, k = 4. Reflection in x-axis, vertical dilation by 2, translation 3 right, translation 4 up.
Step 3 — Transform the vertex.
(0, 0) → (0/1 + 3, −2·0 + 4) = (3, 4).
Step 4 — Find the x-intercepts (set y = 0).
−2(x − 3)² + 4 = 0 ⇒ (x − 3)² = 2 ⇒ x = 3 ± √2.
Step 5 — Find the y-intercept (set x = 0).
y = −2(0 − 3)² + 4 = −18 + 4 = −14, so (0, −14).
Conclusion. A downward parabola with vertex (3, 4), x-intercepts at 3 ± √2, and y-intercept (0, −14). Sketch and label all three.
3. Faded example — fill in the missing steps
Sketch the graph of y = 3 / (x − 2) + 1. Identify its key features. Fill in each blank. 4 marks
Step 1 — Parent function: y = 1/x, with asymptotes x = ____, y = ____, and key points (1, 1) and (−1, −1).
Step 2 — Transformations (write as 3 f(x − 2) + 1): a = ____, b = ____, h = ____, k = ____. Vertical dilation by ____, shift ____ right, shift ____ up.
Step 3 — Asymptotes: VA: x = c/b + h = ____/____ + ____ = ____. HA: y = a·d + k = ____·____ + ____ = ____.
Step 4 — Transform key points: (1, 1) → (____, ____); (−1, −1) → (____, ____).
Conclusion. Sketch a hyperbola with VA x = ____, HA y = ____, passing through (____, ____) and (____, ____).
4. Graduated practice
Foundation — state the vertex/turning point only (4 questions)
| Q | Equation | Vertex / turning point | Parent |
|---|---|---|---|
| 4.1 1 | y = (x − 5)² | ||
| 4.2 1 | y = x² − 3 | ||
| 4.3 1 | y = 2(x + 1)² + 4 | ||
| 4.4 1 | y = |x − 2| + 1 |
Standard — full key-feature analysis (6 questions)
For each function, identify the parent, list transformations, and state vertex/turning point, intercepts, and asymptotes (if any). Show working.
4.5 y = 2(x − 1)² + 3 2 marks
4.6 y = −1/(x + 2) + 3 2 marks
4.7 y = 3√(x − 2) + 1 2 marks
4.8 y = −2|x + 1| + 4 (find x-intercepts as well) 2 marks
4.9 y = 2/(x + 1) − 3 (find asymptotes, x-intercept, y-intercept) 3 marks
4.10 A graph has turning point (2, −3) and x-intercepts at x = −1 and x = 5. Find the new turning point and x-intercepts on y = 2 f(x − 1) + 4. 3 marks
Extension — modelling (2 questions)
4.11 A profit P (thousands of dollars) from selling x hundred items has a maximum profit of $50 000 when 300 items are sold, and breaks even at 100 and 500 items. Find the equation of the model in vertex form. 3 marks
4.12 A ball thrown vertically upward is modelled by h = −5(t − 2)² + 20, with h in metres and t in seconds. Find the initial height, the maximum height, the time of maximum height, and the time the ball hits the ground. 4 marks
5. Self-check the easy 3
Tick once you've verified your method on the foundation row.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Five key features
x-intercepts, y-intercept, turning point(s) / vertex, asymptotes (horizontal, vertical, oblique), domain & range. (End behaviour also acceptable in place of one of the above.)
Q1.2 — Turning point transformation
(x_0 / b + h, a·y_0 + k).
Q1.3 — Asymptote transformations
VA: x = c/b + h. HA: y = a·d + k.
Q3 — Faded example, y = 3/(x − 2) + 1
Step 1: Parent y = 1/x, asymptotes x = 0, y = 0.
Step 2: a = 3, b = 1, h = 2, k = 1. Vertical dilation by 3, shift 2 right, shift 1 up.
Step 3: VA: x = 0/1 + 2 = 2. HA: y = 3·0 + 1 = 1.
Step 4: (1, 1) → (1/1 + 2, 3·1 + 1) = (3, 4). (−1, −1) → (−1/1 + 2, 3·(−1) + 1) = (1, −2).
Conclusion: hyperbola with VA x = 2, HA y = 1, passing through (3, 4) and (1, −2).
Q4.1−4.4 — Vertices/parents
4.1 (5, 0), parabola. 4.2 (0, −3), parabola. 4.3 (−1, 4), parabola. 4.4 (2, 1), absolute-value V-shape.
Q4.5 — y = 2(x − 1)² + 3
Parent: y = x². Vertical dilation 2, right 1, up 3. Vertex (1, 3). y-intercept: y(0) = 2(1) + 3 = 5, so (0, 5). No x-intercepts (vertex is above axis on an upward parabola).
Q4.6 — y = −1/(x + 2) + 3
Parent: y = 1/x. a = −1, h = −2, k = 3. VA: x = −2; HA: y = 3. Reflection in x-axis. y-intercept: y(0) = −1/2 + 3 = 5/2, so (0, 5/2). x-intercept: 0 = −1/(x + 2) + 3 ⇒ 1/(x + 2) = 3 ⇒ x = −5/3, so (−5/3, 0).
Q4.7 — y = 3√(x − 2) + 1
Parent: y = √x. a = 3, h = 2, k = 1. Starting point: (0, 0) → (2, 1). Domain x ≥ 2. Range y ≥ 1. No x-intercepts (curve stays at or above y = 1). y-intercept undefined (x = 0 not in domain).
Q4.8 — y = −2|x + 1| + 4
Parent: y = |x|. a = −2, h = −1, k = 4. Vertex (−1, 4); opens downward. x-intercepts: 0 = −2|x + 1| + 4 ⇒ |x + 1| = 2 ⇒ x + 1 = ±2 ⇒ x = 1 or x = −3. y-intercept: y(0) = −2(1) + 4 = 2, so (0, 2).
Q4.9 — y = 2/(x + 1) − 3
Parent: y = 1/x. a = 2, h = −1, k = −3. VA: x = −1; HA: y = −3. x-intercept: 0 = 2/(x + 1) − 3 ⇒ 2/(x + 1) = 3 ⇒ x + 1 = 2/3 ⇒ x = −1/3, so (−1/3, 0). y-intercept: y(0) = 2/1 − 3 = −1, so (0, −1).
Q4.10 — Image of features on y = 2 f(x − 1) + 4
a = 2, b = 1, h = 1, k = 4. Turning point (2, −3) → (2/1 + 1, 2·(−3) + 4) = (3, −2). x-intercepts (where y = 0 on original) have y_0 = 0 so image y = 2·0 + 4 = 4 — these are not on the new x-axis. Instead the new x-intercepts need a different y_0. To find where the new curve crosses y = 0 we need the original y_0 such that 2 y_0 + 4 = 0 ⇒ y_0 = −2, which is at the turning point. So the new x-intercepts coincide with the new turning point: a single touch at (3, −2) would only occur if the curve touches y = 0 at the vertex; here the vertex is at y = −2 (below 0), so the parabola crosses y = 0 above the vertex — we cannot determine the exact crossings without more info on f. If the question intended "find images of the original x-intercepts on the new graph": originals (−1, 0) and (5, 0) map to (−1 + 1, 2·0 + 4) = (0, 4) and (5 + 1, 2·0 + 4) = (6, 4).
Q4.11 — Profit model
Max profit at x = 3 hundred items, P_max = 50 thousand: vertex (3, 50). Use vertex form P = a(x − 3)² + 50, with a < 0 (downward). At x = 1 (or 5), P = 0: 0 = a(1 − 3)² + 50 ⇒ 4a = −50 ⇒ a = −12.5. P = −12.5(x − 3)² + 50.
Q4.12 — Vertical ball model h = −5(t − 2)² + 20
Initial height: h(0) = −5(4) + 20 = 0 m. Maximum height: h_max = 20 m at vertex. Time of max: t = 2 s. Hits ground when h = 0: −5(t − 2)² + 20 = 0 ⇒ (t − 2)² = 4 ⇒ t = 0 or t = 4, so the ball hits the ground at t = 4 s (rejecting t = 0 as launch).