ECZ 2026 · GCE Paper 2 · Question 7 — Graphs of Functions
Question from ECZ 2026 · GCE Paper 2 · Question 7
(a) The variables x and y are connected by the equation y = x3 - 5x2 - x + 5 as shown in the table. x-2-1012345.2y-21050-9-16q5.2
(a(i)) Calculate the value of q.[1]
Verified working — step 1
q is the y-value at x = 4, so substitute into the equation:(a(ii)) Using a scale of 2cm to represent 1 unit on the x-axis for -2 ≤ x ≤ 6 and 2cm to represent 5 units on the y-axis for -25 ≤ y ≤ 10, draw the graph of y = x3 - 5x2 - x + 5.[3]
Verified working — step 1
Plot the table's points, including q: (-2, -21), (-1, 0),…(a(iii)) Use your graph to solve the equations
(a(iii)(a)) x3 - 5x2 - x + 5 = 0,[2]
Verified working — step 1
x3 - 5x2 - x + 5 = 0 is the curve at y = 0, where it crosses the x-axis.(a(iii)(b)) x3 - 5x2 - x + 5 = 2x - 15.[3]
Verified working — step 1
The equation says curve = line, so draw y = 2x - 15 on the same grid, e.g. through (0, -15) and (5, -5).(b) Find the equation of a tangent to the curve y = x3 - x2 - 15x + 3 at the point (1, -12).[3]
Verified working — step 1
Differentiate: dydx = 3x2 - 2x - 15.The full step-by-step working and final answer are included with Premium — checked against ECZ marking standards.
Original examination question © Examinations Council of Zambia. Worked solution and commentary © G12 Titan — not to be reproduced without permission.
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