Given that x²-12x+k = (x-p)²+5, where k and p are constants. Find the value of k and the value of p.

Mark scheme

Marking points from the official mark scheme. The mark or mark type is shown in brackets.

1. Part (i) M1: For attempting to complete the square x²-12x+k = (x ± 6)² ± ... OR compares coefficients from (x-p)²+5 = x²-2px+p²+5.
2. Part (i) A1: p=6.
3. Part (i) A1: k=41.
4. Part (ii)(a) M1: Sets x²-12x+20=0 and attempts to solve for x.
5. Part (ii)(a) A1: Correct values for x, x=2, x=10.
6. Part (ii)(a) B1: For stating y=0 (or implied by x-intercepts). This mark is often awarded at x=2, x=10.
7. Part (ii)(a) A1: Length of AB=8. (or use distance formula √((10-2)²+(0-0)²))
8. Part (ii)(b) M1: For a correct method to find the x-coordinate of the minimum point. E.g. -b/2a or derivative set to 0. E.g. 2x-12 = 0 ⇒ x=6.
9. Part (ii)(b) A1: Correct coordinates for M=(6,-16).
10. Part (ii)(b) M1: For correct method to find the area of the triangle. Uses their length of AB and their y-coordinate of M. Must be 1/2bh. E.g. Area = 1/2 × 8 × 16.
11. Part (ii)(b) A1: Correct area 64.