Higher June 2024 Paper 1R Q20

EdexcelHigherCurrent spec5 marksCompleting the Square

(a) Express \(\;2x^2 - 11x + 9\;\) in the form \(\;a(x - b)^2 - c\;\) where \(a\), \(b\) and \(c\) are numbers to be found. (3)

The curve C has equation \(\;y = 2(x - 3)^2 - 11(x - 3) + 9\)

The point \(P\) is the minimum point on C

(b) Find the coordinates of \(P\) (2)

Mark scheme (a)

Scheme	Marks
\(2\left(x^2 - \dfrac{11}{2}x\right) + \ldots\) or \(2\left(x^2 - \dfrac{11}{2}x + \ldots\right)\) oe	M1
\(2\left[\left(x - \dfrac{11}{4}\right)^2 - \dfrac{11^2}{4^2}\right] + \ldots\) or \(2\left[\left(x - \dfrac{11}{4}\right)^2 - \dfrac{11^2}{4^2} + \ldots\right]\) oe	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(2\left(x - \dfrac{11}{4}\right)^2 - \dfrac{49}{8}\)	A1
	(3)

M1: for taking out a factor of 2

M1: for correctly completing square

A1: oe, eg \(2(x - 2.75)^2 - 6.125\)

allow \(a = 2\), \(b = \dfrac{11}{4}\) oe, \(c = \dfrac{49}{8}\) oe

if no other marks awarded, award SCB1 for \(2\left(x - \dfrac{11}{4}\right)^2 + \ldots\)

Scheme	Marks
\(ax^2 - 2bax + b^2a - c\)	M1
\(-2ba = -11\) or \(2ba = 11\) and \(b^2a - c = 9\)	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(2\left(x - \dfrac{11}{4}\right)^2 - \dfrac{49}{8}\)	A1
	(3)

M1: for correctly expanding \(a(x - b)^2 - c\) to give \(ax^2 - 2bax + b^2a - c\)

M1: for setting up 2 equations using the coefficient of \(x\) and the numerical term

A1: oe, eg \(2(x - 2.75)^2 - 6.125\)

allow \(a = 2\), \(b = \dfrac{11}{4}\) oe, \(c = \dfrac{49}{8}\) oe

if no other marks awarded, award SCB1 for \(2\left(x - \dfrac{11}{4}\right)^2 + \ldots\)