Unit 3
7 marks · HardThe equation of a curve is y = x^2 − 6x + 8. The straight line y = 2x − 4 is also drawn.
Solve the equation x^2 − 6x + 8 = 2x − 4 algebraically.
Write your solutions as coordinates of the points of intersection between the curve and the line.
On the same axes, sketch the curve and the line, clearly showing the intersection points.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) x^2 - 8x + 12 = 0; x = 6 or x = 2
(b) (2, 0) and (6, 8)
(c) — · must show Parabola opening upwards, crossing x-axis at x = 2 and x = 4, Line with correct gradient and y-intercept, Intersection points clearly marked
Must show: Both curves and lines labelled, Intersection points marked and labelled, Axes labelled
Common mistakes: Not rearranging to zero; Errors with quadratic formula
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Correctly forms equation by equating y = x^2-6x+8 to y=2x-4
M2 (a) · 1 mark · Expands and rearranges to quadratic in standard form
A1 (a) · 1 mark · Solves quadratic: x=6 and x=2
B1 (b) · 1 mark · Substitutes x-values into either equation to get full coordinates (follow-through)
A2 (b) · 1 mark · Final answers: (2,0), (6,8) (follow-through)
D1 (c) · 2 marks · Both the curve and the line correctly sketched; intersection points labelled and axes marked (follow-through)
Worked solution
a) Set x^2 − 6x + 8 = 2x − 4
Factorise: (x-6)(x-2)=0, so x=6 or x=2.
b) For x=2: y=2×2−4=0, so (2,0). For x=6: y=2×6−4=8, so (6,8).
c) Sketch required: Curve y=x^2−6x+8 opens upwards, has roots at x=2 and x=4 (check), but the points of intersection with the line are at x=2 and x=6 as above. Line y=2x−4 crosses y-axis at −4.