IGCSE Algebra Completing the Square Past Paper Questions

Higher June 2025 Paper 2 Q25

EdexcelCurrent spec6 marksCompleting the Square Drawing & Using Graphs

(a) Write \(28 + 24x - 6x^2\) in the form \(a - b(x - c)^2\) where \(a\), \(b\) and \(c\) are integers. (3)

(b) On the axes below, sketch the graph of \(y = 28 + 24x - 6x^2\)
Show clearly the coordinates of the turning point and the coordinates of the point of intersection of the graph with the \(y\)-axis. (3)

Mark scheme (a)

Scheme	Marks
\(\pm 6\left(x \pm \dfrac{4}{2}\right)^2\ldots\ldots\) or \(\pm 6(x \pm 2)^2\ldots\ldots\) or \(\pm 6\left[(x \pm 2)^2\ldots\ldots\right]\) or \(\pm 6\left[\left(x \pm \dfrac{4}{2}\right)^2\ldots\ldots\right]\) or \(\pm 6\left(x \pm \dfrac{24}{2 \times -6}\right)^2\ldots\)	M1
\(-6\left[\left(x - \dfrac{4}{2}\right)^2 - \left(\dfrac{4}{2}\right)^2\right]\ldots\ldots\) or \(-6\left[(x - 2)^2 - 2^2\right]\ldots\ldots\) or \(-6\left[\left(x - \dfrac{4}{2}\right)^2 - \left(\dfrac{4}{2}\right)^2\ldots\ldots\right]\) or \(-6\left[(x - 2)^2 - 2^2\ldots\ldots\right]\) or \(-6\left(x + \dfrac{24}{2 \times -6}\right)^2 - \dfrac{24^2}{4 \times -6}\ldots\ldots\)	M1
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: \(52 - 6(x - 2)^2\)	A1
	(3)

Notes

M1: for a start to completing the square

correct substitution into \(a\left(x + \dfrac{b}{2a}\right)^2 + \ldots\) from the formula \(a\left(x + \dfrac{b}{2a}\right)^2 - \dfrac{(b)^2}{4a} + c\)

M1: for correctly completing the square but terms do not need to be simplified and 28 may or may not be present

correct simplification of the first two parts of \(a\left(x + \dfrac{b}{2a}\right)^2 - \dfrac{(b)^2}{4a}(+c)\)

NB: Please refer to ALT mark scheme after (b) for comparison of coefficients method

A1: oe eg \(-6(x - 2)^2 + 52\)

25(a) ALT

Scheme	Marks
\(-bx^2 + 2bcx - bc^2 + a\) and \(b = 6\) or \(b = -6\)	M1
\(2bc = 24\) or \(-bc^2 + a = 28\)	M1
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: \(52 - 6(x - 2)^2\)	A1

Notes

M1: for multiplying out \(a - b(x - c)^2\) and \(b = 6\) or \(b = -6\)

M1: for equating coefficients

A1: oe eg \(-6(x - 2)^2 + 52\)

Sketch of an n-shaped curve with turning point marked (2, 52) — Mark scheme (b)

Sketch of an n-shaped curve crossing the y-axis at (0, 28) — Mark scheme (b)

Scheme	Marks
‘∩’ or ‘∪’ shaped symmetrical quadratic curve	B1
Answer: Turning point marked as (2, 52)	B1
Answer: Intersection with \(y\)-axis marked as (0, 28) or crossing at 28 marked	B1
	(3)
	(6 marks)

Higher November 2024 Paper 1 Q25

EdexcelCurrent spec4 marksCompleting the Square Functions

25 The function f is such that \(\mathrm{f}(x) = 2x^2 - 24x + 7\) where \(x \geqslant 6\)

Find the inverse function \(\mathrm{f}^{-1}(x)\)

(4)

Mark scheme

Scheme	Marks
\((y =)\;2(x^2 - 12x) + 7\) or \((y =)\;2\left(x^2 - 12x + \dfrac{7}{2}\right)\) or \(\dfrac{y - 7}{2} = x^2 - 12x\) or \((x =)\;2(y^2 - 12y) + 7\) or \((x =)\;2\left(y^2 - 12y + \dfrac{7}{2}\right)\) or \(\dfrac{x - 7}{2} = y^2 - 12y\)	M1
eg \((y =)\;2\left((x - 6)^2 - 6^2\right) + 7\) or \((y =)\;2\left((x - 6)^2 - 6^2 + \dfrac{7}{2}\right)\) or \((y =)\;2(x - 6)^2 - 65\) oe or \(\dfrac{y - 7}{2} = (x - 6)^2 - 6^2\) oe or eg \((x =)\;2\left((y - 6)^2 - 6^2\right) + 7\) or \((x =)\;2\left((y - 6)^2 - 6^2 + \dfrac{7}{2}\right)\) or \((x =)\;2(y - 6)^2 - 65\) oe or \(\dfrac{x - 7}{2} = (y - 6)^2 - 6^2\) oe	M1
\((x - 6)^2 = \dfrac{y + 65}{2}\) oe or \((x - 6)^2 = \dfrac{y - 7}{2} + 6^2\) oe or \((y - 6)^2 = \dfrac{x + 65}{2}\) oe or \((y - 6)^2 = \dfrac{x - 7}{2} + 6^2\) oe	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(6 + \sqrt{\dfrac{x + 65}{2}}\)	A1
	(4)
	(4 marks)

Notes

M1: for a correct first step in order to complete the square

For each method mark the function must be correct

M1: dep on M1

A1: oe eg \(6 + \sqrt{\dfrac{x - 7}{2} + 36}\)

Must be in terms of \(x\)

M3A0 for \(6 \pm \sqrt{\dfrac{x + 65}{2}}\) or \(6 \pm \sqrt{\dfrac{y + 65}{2}}\)

or \(6 + \sqrt{\dfrac{y + 65}{2}}\)

Note: Allow candidates to swap \(x\) and \(y\) when finding the inverse

Mark scheme (ALT)

Scheme	Marks
\(2x^2 - 24x + 7 - y\;(= 0)\)	M1
\((x =)\;\dfrac{24 \pm \sqrt{576 - 8(7 - y)}}{4}\) or \((x =)\;\dfrac{24 + \sqrt{576 - 8(7 - y)}}{4}\) or \(2\left((x - 6)^2 - 6^2\right) + 7 - y\;(= 0)\) or \(2\left((x - 6)^2 - 6^2 + \dfrac{7}{2}\right) - y\;(= 0)\)	M1
\((x =)\;6 \pm \sqrt{\dfrac{y + 65}{2}}\) or \((x - 6)^2 = \dfrac{y + 65}{2}\)	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(6 + \sqrt{\dfrac{x + 65}{2}}\)	A1

Notes

M1: for a correct first step

M1: dep on M1

A1: oe eg \(6 + \sqrt{\dfrac{x - 7}{2} + 36}\)

Must be in terms of \(x\)

M3A0 for \(6 \pm \sqrt{\dfrac{x + 65}{2}}\) or \(6 \pm \sqrt{\dfrac{y + 65}{2}}\)

or \(6 + \sqrt{\dfrac{y + 65}{2}}\)

Note: Allow candidates to swap \(x\) and \(y\) when finding the inverse