WMA11/01 Oct 2019, Question 10: Factorising Quadratic Expressions
Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 10 marks
The question, in short
Use the cubic f(x) = (2x + 5)(x − 3)² and its sketch to find where it is non-positive, expand it, locate point P and its gradient, then translate the curve 2 units right.
Summarised in our own words. Use the official paper for the exact wording and any diagrams.
Part (a) · 2 marks
State the x values satisfying f(x) ≤ 0, using the sketch and factorised form.
- 1f(x) = 0 at x = −5/2 (crosses) and x = 3 (touches, since (x − 3)² ≥ 0)
- 2(x − 3)² ≥ 0, so f(x) ≤ 0 needs 2x + 5 ≤ 0 or x = 3
- 3x ≤ −5/2 or x = 3
Answerx ≤ −5/2 or x = 3
Part (b) · 3 marks
Expand f(x) into the form ax³ + bx² + cx + d with integer coefficients a, b, c, d.
- 1(x − 3)² = x² − 6x + 9
- 2(2x + 5)(x² − 6x + 9) = 2x³ − 12x² + 18x + 5x² − 30x + 45
- 3= 2x³ − 7x² − 12x + 45
Answerf(x) = 2x³ − 7x² − 12x + 45
Part (ci) · 1 mark
Give the coordinates of the point P where the curve meets the y-axis.
- 1At x = 0, f(0) = 45
- 2P = (0, 45)
AnswerP = (0, 45)
Part (cii) · 1 mark
Find the gradient of the curve at P.
- 1f′(x) = 6x² − 14x − 12
- 2At x = 0, gradient = −12
Answer−12
Part (di) · 2 marks
Write g(x), the curve translated 2 units in the positive x direction, in simplified factorised form.
- 1Translation 2 units right: g(x) = f(x − 2)
- 2g(x) = (2(x − 2) + 5)((x − 2) − 3)²
- 3g(x) = (2x + 1)(x − 5)²
Answerg(x) = (2x + 1)(x − 5)²
Part (dii) · 1 mark
State the y intercept of y = g(x).
- 1g(0) = (1)(−5)²
- 2y-intercept = 25
Answer25 (the point (0, 25))
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