June 2023 Paper 3 Q5

5 In this question you must show detailed reasoning.

This question is about the curve \(y = x^3 - 5x^2 + 6x\).

(a) Find the equation of the tangent, T, to the curve at the point \((0, 0)\). [3]

(b) Find the equation of the normal, N, to the curve at the point \((1, 2)\). [3]

Mark scheme (a)

Scheme	Marks	AO
DR \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 10x + 6\)	M1	1.1
When \(x = 0\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6\)	M1	1.1
Tangent goes through origin so equation is \(y = 6x\) cao	A1	1.1
	[3]

M1: At least two terms correct

M1: FT their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

A1: Need reasoning eg \(y - 0 = 6(x - 0)\) or use of \(y = mx + c\)
\(y = 6x\) implies previous M mark

Mark scheme (b)

Scheme	Marks	AO
DR When \(x = 1\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -1\)	M1	1.1
Gradient of normal is 1	M1	1.1
[\((y-2) = (x-1)\) so] \(y = x + 1\)	A1	1.1
	[3]

M1: FT their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) from (a)

M1: FT negative reciprocal of their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).

A1: oe but constant terms should be collected