This set has twelve original questions, ordered from easier to harder, covering all five lessons in graphs and transformations. Questions 1 to 4 cover scale, plotting and lines, 5 to 8 cover translation and reflection, and 9 to 12 mix skills and intersections.
Attempt each question on paper, with a ruler where a drawing helps, and write your working as you would in an exam. Only then open the answer. Mark the ones you got wrong and use the routing list at the end.
Questions
1. A quantity runs from 0 to 320 and you have a 16 cm axis. Suggest a scale and say how long the axis will be.
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Try 1 cm = 20 units: 320 ÷ 20 = 16 cm, so it fits exactly and each centimetre is easy to read. A scale of 1 cm = 10 units would need 32 cm, which is too long.
1 cm = 20 units, axis length 16 cm.
2. P(−4, 3), Q(2, 3) and R(2, −5) are three corners of a rectangle PQRS. Find S and the area.
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PQ is horizontal with length 2 − (−4) = 6. QR is vertical with length 3 − (−5) = 8. S shares x = −4 with P and y = −5 with R, so S = (−4, −5).
Area = 6 × 8 = 48 square units.
3. A line passes through (−2, 7) and (4, −5). Find its gradient and its equation.
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Gradient = (−5 − 7) ÷ (4 − (−2)) = −12 ÷ 6 = −2.
Use y − 7 = −2(x + 2), so y = −2x − 4 + 7 = −2x + 3.
Check with (4, −5): −2 × 4 + 3 = −5.
Gradient −2, equation y = −2x + 3.
4. A gym charges a joining fee plus a fee per class. The graph of cost C (RM) against classes n passes through (0, 45) and (5, 105). Find the equation and the cost of 9 classes, and explain the gradient and intercept.
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Gradient = (105 − 45) ÷ 5 = 60 ÷ 5 = 12. Intercept is 45. So C = 12n + 45.
For 9 classes: 12 × 9 + 45 = 108 + 45 = RM153.
The gradient, 12, means each extra class costs RM12. The intercept, 45, is the cost with no classes, the joining fee of RM45.
5. The height h (cm) of water in a leaking bucket after t hours is h = 60 − 4t. Interpret both numbers and find when the bucket is empty.
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The gradient −4 means the water level drops by 4 cm every hour. The intercept 60 means the starting height is 60 cm.
Empty when 60 − 4t = 0, so t = 60 ÷ 4 = 15 hours.
Check: 60 − 4 × 15 = 0.
6. The graph of y = x² is translated by the vector (−2, 5). Write the new equation and the vertex. Where does the point (3, 9) go?
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Moving 2 left replaces x with (x + 2). Moving 5 up adds 5. So y = (x + 2)² + 5, with vertex (−2, 5).
The point (3, 9) becomes (3 − 2, 9 + 5) = (1, 14).
Check: (1 + 2)² + 5 = 9 + 5 = 14.
7. The line y = 3x − 2 is translated (a) by the vector (0, −4) and (b) by the vector (2, 0). Find each new equation.
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(a) Subtract 4 from y: y = 3x − 2 − 4 = y = 3x − 6.
(b) Replace x with (x − 2): y = 3(x − 2) − 2 = 3x − 6 − 2 = y = 3x − 8.
Check (b): the point (0, −2) moves to (2, −2), and 3 × 2 − 8 = −2.
Check (a): (0, −2) moves to (0, −6), and 3 × 0 − 6 = −6.
8. The curve y = x² − 6x + 5 has vertex (3, −4) and roots 1 and 5. Find the equation and vertex of its reflection (a) in the x-axis and (b) in the y-axis.
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(a) Negate the whole function: y = −x² + 6x − 5. The vertex is (3, 4) and the roots stay at 1 and 5.
(b) Replace x with −x: y = x² + 6x + 5. The vertex is (−3, −4) and the roots are −1 and −5, since x² + 6x + 5 = (x + 1)(x + 5).
Check (a) at x = 3: −9 + 18 − 5 = 4. Check (b) at x = −3: 9 − 18 + 5 = −4.
9. The point (−3, 8) is on y = f(x). Give the corresponding point on (a) y = f(x) + 2, (b) y = f(x + 2), (c) y = −f(x), (d) y = f(−x).
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(a) Add 2 to y: (−3, 10).
(b) Move 2 left: (−5, 8). Check: f(−5 + 2) = f(−3) = 8.
(c) Negate y: (−3, −8).
(d) Negate x: (3, 8). Check: f(−3) = 8.
10. Use the graphs of y = x² and y = x + 5 to solve x² = x + 5, giving answers to 1 decimal place. Explain how you would check them.
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The crossings have x-values near 2.8 and −1.8. The formula gives x = (1 ± √21) ÷ 2, which is about 2.79 and −1.79.
Check: at x = 2.8, x² = 7.84 and x + 5 = 7.8. At x = −1.8, x² = 3.24 and x + 5 = 3.2. Both differences are 0.04, small enough for a graph reading.
11. The line y = 2x − 1 is reflected in the x-axis and then translated 6 units up. Find the final equation and check it with a point.
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Reflect in the x-axis: y = −(2x − 1) = −2x + 1. Translate up 6: y = −2x + 1 + 6 = y = −2x + 7.
Check: (1, 1) is on the original. Reflecting gives (1, −1). Translating gives (1, 5). Then −2 × 1 + 7 = 5.
12. The graphs of y = x² − 1 and y = 3 − x intersect. Which equation do their intersections solve, and what are the solutions to 1 decimal place?
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Set x² − 1 = 3 − x, so x² + x − 4 = 0. The solutions are x = (−1 ± √17) ÷ 2, which are about 1.56 and −2.56.
To 1 decimal place: x ≈ 1.6 and x ≈ −2.6.
Check: at x = 1.56, x² − 1 = 1.4336 and 3 − x = 1.44. The two sides agree closely.
If you got these wrong
- Scale or plotting (questions 1 and 2): review plotting coordinates using an appropriate scale.
- Gradient, equation or meaning (questions 3, 4 and 5): review interpreting gradient and intercept in a context.
- Direction of a shift (questions 6, 7 and parts of 9 and 11): review translating a simple graph.
- Which axis flips (questions 8, 9 and 11): review the effect of reflecting a graph.
- Intersections (questions 10 and 12): review using intersections to approximate a solution.
Record your repeated slips in the mistake log and retest queue, and use the quadratic structure explorer or the graph model explorer to see a change before you calculate it.
If you still flip the wrong axis or shift the wrong way, online one-to-one Mathematics tuition gives you a teacher to work through your method with you.