# IB Math Analysis & Approaches Questionbank-Topic: SL 2.6 quadratic function SL Paper 1

## Question

Let $$f$$ be a quadratic function. Part of the graph of $$f$$ is shown below.

The vertex is at P($$4$$, $$2$$) and the y-intercept is at Q($$0$$, $$6$$) .

Write down the equation of the axis of symmetry.

[1]
a.

The function f can be written in the form $$f(x) = a{(x – h)^2} + k$$ .

Write down the value of h and of k .

[2]
b.

The function f can be written in the form $$f(x) = a{(x – h)^2} + k$$ .

Find a .

[3]
c.

## Markscheme

$$x = 4$$ (must be an equation)Â Â Â Â  A1Â Â Â Â  N1

[1 mark]

a.

$$h = 4$$ , $$k = 2$$Â Â Â Â  A1A1Â Â Â Â  N2

[2 marks]

b.

attempt to substitute coordinates of any point on the graph into fÂ Â Â Â  (M1)

e.g. $$f(0) = 6$$ , $$6 = a{(0 – 4)^2} + 2$$ , $$f(4) = 2$$

correct equation (do not accept an equation that results from $$f(4) = 2$$ )Â Â Â Â  (A1)

e.g. $$6 = a{( – 4)^2} + 2$$ , $$6 = 16a + 2$$

$$a = \frac{4}{{16}}\left( { = \frac{1}{4}} \right)$$Â Â Â Â  A1 Â  Â  N2

[3 marks]

c.

## Question

LetÂ $$f(x) = \sin x + \frac{1}{2}{x^2} – 2x$$ , forÂ $$0 \le x \le \pi$$ .

Let $$g$$ be a quadratic function such that $$g(0) = 5$$ . The line $$x = 2$$ is the axis of symmetry of the graph of $$g$$ .

The function $$g$$ can be expressed in the form $$g(x) = a{(x – h)^2} + 3$$ .

Find $$f'(x)$$ .

[3]
a.

FindÂ $$g(4)$$ .

[3]
b.

(i)Â Â Â Â  Write down the value of $$h$$ .

(ii)Â Â Â Â  Find the value of $$a$$ .

[4]
c.

Find the value of $$x$$ for which the tangent to the graph of $$f$$ is parallel to the tangent to the graph of $$g$$ .

[6]
d.

## Markscheme

$$f'(x) = \cos x + x – 2$$Â Â Â Â  A1A1A1Â Â Â Â  N3

Note: Award A1 for each term.

[3 marks]

a.

recognizing $$g(0) = 5$$Â gives the point ($$0$$, $$5$$)Â Â Â  Â (R1)

recognize symmetryÂ Â Â Â  (M1)

eg vertex, sketch

$$g(4) = 5$$Â Â Â Â  A1Â Â Â Â  N3

[3 marks]

b.

(i)Â Â Â Â  $$h = 2$$Â Â Â Â  A1 N1

(ii)Â Â Â Â  substituting intoÂ $$g(x) = a{(x – 2)^2} + 3$$Â (not the vertex)Â Â Â Â  (M1)

eg Â  $$5 = a{(0 – 2)^2} + 3$$ , $$5 = a{(4 – 2)^2} + 3$$

working towards solutionÂ Â Â Â  (A1)

eg Â  $$5 = 4a + 3$$ , $$4a = 2$$

$$a = \frac{1}{2}$$Â Â Â Â  A1Â Â Â Â  N2

[4 marks]

c.

$$g(x) = \frac{1}{2}{(x – 2)^2} + 3 = \frac{1}{2}{x^2} – 2x + 5$$

correct derivative of $$g$$Â Â Â Â  A1A1

eg Â  $$2 \times \frac{1}{2}(x – 2)$$ , $$x – 2$$

evidence of equating both derivativesÂ Â Â Â  (M1)

eg Â  $$f’ = g’$$

correct equationÂ Â Â Â  (A1)

eg Â  $$\cos x + x – 2 = x – 2$$

working towards a solutionÂ Â Â Â  (A1)

egÂ Â  $$\cos x = 0$$Â , combining like terms

$$x = \frac{\pi }{2}$$Â Â Â  A1Â Â Â Â  N0

Note: Do not award final A1 if additional values are given.

[6 marks]

d.

## Question

Let $$f(x) = a{(x – h)^2} + k$$. The vertex of the graph of $$f$$ is at $$(2, 3)$$ and the graph passesÂ through $$(1, 7)$$.

Write down the value ofÂ $$h$$Â and ofÂ $$k$$.

[2]
a.

Find the value of $$a$$.

[3]
b.

## Markscheme

$$h = 2,{\text{ }}k = 3$$ Â  Â Â A1A1 Â  Â  N2

[2 marks]

a.

attempt to substitute $$(1,7)$$ in any order into theirÂ $$f(x)$$Â  Â  Â (M1)

eg Â  Â Â $$7 = a{(1 – 2)^2} + 3{\text{, }}7 = a{(1 – 3)^2} + 2{\text{, }}1 = a{(7 – 2)^2} + 3$$

correct equation Â  Â  (A1)

eg Â  Â Â $$7 = a + 3$$

a = 4 Â  Â Â A1 Â  Â  N2

[3 marks]

b.

[N/A]

a.

[N/A]

b.

## Question

Let $$f(x) = 3{x^2} – 6x + p$$. The equation $$f(x) = 0$$ has two equal roots.

Write down the value of the discriminant.

[2]
a(i).

Hence, show that $$p = 3$$.

[1]
a(ii).

The graph of $$f$$has its vertex on the $$x$$-axis.

Find the coordinates of the vertex of the graph of $$f$$.

[4]
b.

The graph of $$f$$ has its vertex on the $$x$$-axis.

Write down the solution of $$f(x) = 0$$.

[1]
c.

The graph of $$f$$Â has its vertex on the $$x$$-axis.

The function can be written in the form $$f(x) = a{(x – h)^2} + k$$. Write down the value of $$a$$.

[1]
d(i).

The graph of $$f$$ has its vertex on the $$x$$-axis.

The function can be written in the form $$f(x) = a{(x – h)^2} + k$$. Write down the value of $$h$$.

[1]
d(ii).

The graph of $$f$$ has its vertex on the $$x$$-axis.

The function can be written in the form $$f(x) = a{(x – h)^2} + k$$. Write down the value of $$k$$.

[1]
d(iii).

The graph of $$f$$ has its vertex on the $$x$$-axis.

The graph of a function $$g$$ is obtained from the graph of $$f$$ by a reflection of $$f$$ in the $$x$$-axis, followed by a translation by the vector $$\left( \begin{array}{c}0\\6\end{array} \right)$$. Find $$g$$, giving your answer in the form $$g(x) = A{x^2} + Bx + C$$.

[4]
e.

## Markscheme

correct value $$0$$, or $$36 – 12p$$ Â  Â  A2 Â  Â  N2

[2 marks]

a(i).

correct equation which clearly leads to $$p = 3$$ Â  Â  A1

eg Â  Â  $$36 – 12p = 0,{\text{ }}36 = 12p$$

$$p = 3$$ Â  Â  AG Â  Â  N0

[1 mark]

a(ii).

METHOD 1

valid approach Â  Â  (M1)

eg Â  Â  $$x =Â – \frac{b}{{2a}}$$

correct working Â  Â  A1

eg Â  Â  $$– \frac{{( – 6)}}{{2(3)}},{\text{ }}x = \frac{6}{6}$$

correct answers Â  Â  A1A1 Â  Â  N2

eg Â  Â  $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$

METHOD 2

valid approach Â  Â  (M1)

eg Â  Â  $$f(x) = 0$$, factorisation, completing the square

correct working Â  Â  A1

eg Â  Â  $${x^2} – 2x + 1 = 0,{\text{ }}(3x – 3)(x – 1),{\text{ }}f(x) = 3{(x – 1)^2}$$

correct answers Â  Â  A1A1 Â  Â  N2

eg Â  Â  $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$

METHOD 3

valid approach using derivative Â  Â  (M1)

eg Â  Â  $$f'(x) = 0,{\text{ }}6x – 6$$

correct equation Â  Â  A1

eg Â  Â  $$6x – 6 = 0$$

correct answers Â  Â  A1A1 Â  Â  N2

eg Â  Â  $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$

[4 marks]

b.

$$x = 1$$ Â  Â  A1 Â  Â  N1

[1 mark]

c.

$$a = 3$$ Â  Â  A1 Â  Â  N1

[1 mark]

d(i).

$$h = 1$$ Â  Â  A1 Â  Â  N1

[1 mark]

d(ii).

$$k = 0$$ Â  Â  A1 Â  Â  N1

[1 mark]

d(iii).

attempt to apply vertical reflection Â  Â  (M1)

eg Â  Â Â $$– f(x),{\text{ }} – 3{(x – 1)^2}$$, sketch

attempt to apply vertical shift 6 units up Â  Â  (M1)

eg Â  Â Â $$– f(x) + 6$$, vertex $$(1, 6)$$

transformations performed correctly (in correct order) Â  Â  (A1)

eg Â  Â Â $$– 3{(x – 1)^2} + 6,{\text{ }} – 3{x^2} + 6x – 3 + 6$$

$$g(x) =Â – 3{x^2} + 6x + 3$$ Â  Â  A1 Â  Â  N3

[4 marks]

e.

## Question

Consider $$f(x) = {x^2} + qx + r$$. The graph of $$f$$ has a minimum value when $$x = Â – 1.5$$.

The distance between the two zeros of $$f$$ is 9.

Show that the two zeros are 3 and $$– 6$$.

[2]
a.

Find the value of $$q$$ and of $$r$$.

[4]
b.

## Markscheme

recognition that the $$x$$-coordinate of the vertex is $$– 1.5$$Â (seen anywhere) Â  Â  (M1)

eg$$\,\,\,\,\,$$axis of symmetry is $$– 1.5$$, sketch,Â $$f'( – 1.5) = 0$$

correct working to find the zeroes Â  Â  A1

eg$$\,\,\,\,\,$$$$– 1.5 \pm 4.5$$

$$x = Â – 6$$ and $$x = 3$$Â Â  Â  AG Â  Â  N0

[2 marks]

a.

METHOD 1 (using factors)

attempt to write factors Â  Â  (M1)

eg$$\,\,\,\,\,$$$$(x – 6)(x + 3)$$

correct factors Â  Â  A1

eg$$\,\,\,\,\,$$$$(x – 3)(x + 6)$$

$$q = 3,{\text{ }}r = Â – 18$$ Â  Â A1A1 Â  Â  N3

METHOD 2 (using derivative or vertex)

valid approach to find $$q$$ Â  Â  (M1)

eg$$\,\,\,\,\,$$$$f'( – 1.5) = 0,{\text{ }} – \frac{q}{{2a}} = Â – 1.5$$

$$q = 3$$ Â  Â A1

correct substitution Â  Â  A1

eg$$\,\,\,\,\,$$$${3^2} + 3(3) + r = 0,{\text{ }}{( – 6)^2} + 3( – 6) + r = 0$$

$$r = Â – 18$$ Â  Â A1

$$q = 3,{\text{ }}r = Â – 18$$ Â  Â N3

METHOD 3 (solving simultaneously)

valid approach setting up system of two equations Â  Â  (M1)

eg$$\,\,\,\,\,$$$$9 + 3q + r = 0,{\text{ }}36 – 6q + r = 0$$

one correct value

eg$$\,\,\,\,\,$$$$q = 3,{\text{ }}r = Â – 18$$Â Â  Â  A1

correct substitution Â  Â  A1

eg$$\,\,\,\,\,$$$${3^2} + 3(3) + r = 0,{\text{ }}{( – 6)^2} + 3( – 6) + r = 0,{\text{ }}{3^2} + 3q – 18 = 0,{\text{ }}36 – 6q – 18 = 0$$

second correct value Â  Â  A1

eg$$\,\,\,\,\,$$$$q = 3,{\text{ }}r = Â – 18$$

$$q = 3,{\text{ }}r = Â – 18$$ Â  Â N3

[4 marks]

b.

## Question

Let f(x) = ax2 âˆ’ 4x âˆ’ c. A horizontal line, L , intersects the graph of f at x = âˆ’1 and x = 3.

The equation of the axis of symmetry is x = p. Find p.

[2]
a.i.

Hence, show that a = 2.

[2]
a.ii.

The equation of L is y = 5 . Find the value of c.

[3]
b.

## Markscheme

METHOD 1 (using symmetry to findÂ p)

valid approachÂ  Â  Â  (M1)

egÂ Â $$\frac{{ – 1 + 3}}{2}$$,Â

p = 1Â  Â  Â A1 N2

Note: Award no marks if they work backwards by substituting a = 2 intoÂ $$– \frac{b}{{2a}}$$ to find p.

Do not acceptÂ $$p = \frac{2}{a}$$.

METHOD 2 (calculating a first)
(i) & (ii) valid approach to calculate aÂ  Â  Â  M1

egÂ  Â a + 4Â âˆ’ c = a(32) âˆ’ 4(3)Â âˆ’ c,Â Â f(âˆ’1) = f(3)

correct workingÂ  Â  Â  A1

egÂ  Â 8a = 16

a = 2Â  Â  Â  AG N0

valid approach to find pÂ  Â  Â Â (M1)

egÂ  Â $$– \frac{b}{{2a}},\,\,\,\frac{4}{{2\left( 2 \right)}}$$

p = 1Â  Â  Â  A1 N2

[2 marks]

a.i.

METHOD 1

valid approachÂ  Â  Â  Â M1

egÂ  $$– \frac{b}{{2a}},\,\,\,\frac{4}{{2a}}$$Â (might be seen in (i)), f’â€‰(1) = 0

correct equationÂ  Â  Â A1

egÂ  $$\frac{4}{{2a}}$$ = 1, 2a(1)Â âˆ’ 4 = 0

a = 2Â  Â  Â  AG N0

METHOD 2 (calculating a first)
(i) & (ii) valid approach to calculate aÂ  Â  Â  M1

egÂ  Â a + 4Â âˆ’ c = a(32) âˆ’ 4(3)Â âˆ’ c,Â Â f(âˆ’1) = f(3)

correct workingÂ  Â  Â  A1

egÂ  Â 8a = 16

a = 2Â  Â  Â  AG N0

[2 marks]

a.ii.

valid approachÂ  Â  Â  (M1)
egÂ  Â f(âˆ’1) = 5, f(3) =5

correct workingÂ  Â  Â  Â (A1)
egÂ  Â 2 + 4Â âˆ’ c = 5, 18Â âˆ’ 12Â âˆ’ c = 5

cÂ = 1Â  Â  Â A1 N2

[3 marks]

b.

## Question

Let $$f\left( x \right) = p{x^2} + qx – 4p$$, where pÂ â‰  0. FindÂ Find the number of roots for the equation $$f\left( x \right) = 0$$.

## Markscheme

METHOD 1

evidence of discriminantÂ  Â  Â  (M1)
egÂ Â $${b^2} – 4ac,\,\,\Delta$$

correct substitution into discriminantÂ  Â  Â  (A1)
egÂ Â $${q^2} – 4p\left( { – 4p} \right)$$

correct discriminantÂ  Â  Â  Â A1
egÂ Â $${q^2}Â + 16{p^2}$$

$$16{p^2} > 0\,\,\,\,\left( {{\text{accept}}\,\,{p^2} > 0} \right)$$Â  Â  Â A1

$${q^2} \geqslant 0\,\,\,\,\left( {{\text{do not accept}}\,\,{q^2} > 0} \right)$$Â  Â  Â A1

$${q^2} + 16{p^2} > 0$$Â  Â  Â  A1

$$f$$ has 2 rootsÂ  Â  Â A1 N0

METHOD 2

y-intercept = âˆ’4p (seen anywhere)Â  Â  Â  A1

if p is positive, then the y-intercept will be negativeÂ  Â  Â  A1

an upward-opening parabola with a negative y-interceptÂ  Â  Â  R1
egÂ  sketch that must indicate p > 0.

if p is negative, then the y-intercept will be positiveÂ  Â  Â  A1

a downward-opening parabola with a positive y-interceptÂ  Â  Â  R1
egÂ  sketch that must indicate pÂ >Â 0.

$$f$$ has 2 rootsÂ  Â  Â A2 N0

[7 marks]

## Question

LetÂ $$f(x) = 3{(x + 1)^2} – 12$$ .

Show that $$f(x) = 3{x^2} + 6x – 9$$ .

[2]
a.

For the graph of f

(i)Â Â Â Â  write down the coordinates of the vertex;

(ii)Â Â Â  write down the y-intercept;

(iii)Â Â  find both x-intercepts.

[7]
b(i), (ii) and (iii).

Hence sketch the graph of f .

[3]
c.

Let $$g(x) = {x^2}$$ . The graph of f may be obtained from the graph of g by the following two transformations

a stretch of scale factor t in the y-direction,

followed by a translation of $$\left( \begin{array}{l} p\\ q \end{array} \right)$$ .

Write down $$\left( \begin{array}{l} p\\ q \end{array} \right)$$ and the value of t .

[3]
d.

## Markscheme

$$f(x) = 3({x^2} + 2x + 1) – 12$$Â Â Â Â Â A1

$$= 3{x^2} + 6x + 3 – 12$$Â Â Â Â Â A1

$$= 3{x^2} + 6x – 9$$Â Â Â Â Â AGÂ Â Â Â  N0

[2 marks]

a.

(i) vertex is $$( – 1, – 12)$$Â Â Â Â Â A1A1Â Â Â Â  N2

(ii) $$y = – 9$$ , or $$(0, – 9)$$Â Â Â  Â A1 Â  Â  N1

(iii) evidence of solving $$f(x) = 0$$Â Â Â Â Â M1

e.g. factorizing, formula

correct workingÂ Â Â Â  A1

e.g. $$3(x + 3)(x – 1) = 0$$ , $$x = \frac{{ – 6 \pm \sqrt {36 + 108} }}{6}$$

$$x = – 3$$ , $$x = 1$$ , or $$( – 3{\text{, }}0){\text{, }}(1{\text{, }}0)$$Â Â Â Â Â A1A1 Â  Â  N2

[7 marks]

b(i), (ii) and (iii).

Â Â Â Â  A1A1A1 Â  Â  N3

Note: Award A1 for a parabola opening upward, A1 for vertex in approximately correct position, A1 for intercepts in approximately correct positions. Scale and labelling not required.

[3 marks]

c.

$$\left( \begin{array}{l} p\\ q \end{array} \right) = \left( \begin{array}{l} – 1\\ – 12 \end{array} \right)$$ , $$t = 3$$Â Â Â Â  A1A1A1Â Â Â Â  N3

[3 marks]

d.

## Question

Let $$f(x) = 3{(x + 1)^2} – 12$$ .

Show that $$f(x) = 3{x^2} + 6x – 9$$ .

[2]
a.

For the graph of f

(i)Â Â Â Â  write down the coordinates of the vertex;

(ii)Â Â Â  write down the equation of the axis of symmetry;

(iii) Â  write down the y-intercept;

(iv)Â Â  find both x-intercepts.

[8]
b(i), (ii), (iii) and (iv).

Hence sketch the graph of f .

[2]
c.

Let $$g(x) = {x^2}$$ . The graph of f may be obtained from the graph of g by the two transformations:

a stretch of scale factor t in the y-direction

followed by a translation of $$\left( {\begin{array}{*{20}{c}} p\\ q \end{array}} \right)$$ .

Find $$\left( {\begin{array}{*{20}{c}} p\\ q \end{array}} \right)$$Â and the value of t.

[3]
d.

## Markscheme

$$f(x) = 3({x^2} + 2x + 1) – 12$$Â Â Â  Â A1

$$= 3{x^2} + 6x + 3 – 12$$Â Â Â Â Â A1

$$= 3{x^2} + 6x – 9$$Â Â Â Â Â AGÂ Â Â Â  N0

[2 marks]

a.

(i) vertex is $$( – 1{\text{, }} – 12)$$ Â Â Â  A1A1Â Â Â Â  N2

(ii) $$x = – 1$$ (must be an equation)Â Â Â Â  A1 Â  Â  N1

(iii) $$(0{\text{, }} – 9)$$ Â Â Â  A1Â Â Â Â  N1

(iv) evidence of solving $$f(x) = 0$$Â Â Â  Â (M1)

e.g. factorizing, formula,

correct workingÂ Â Â Â  A1

e.g.Â $$3(x + 3)(x – 1) = 0$$ , $$x = \frac{{ – 6 \pm \sqrt {36 + 108} }}{6}$$

$$( – 3{\text{, }}0)$$, $$(1{\text{, }}0)$$ Â Â Â  A1A1 Â  Â  N1N1

[8 marks]

b(i), (ii), (iii) and (iv).

Â Â Â Â  A1A1Â Â Â Â  N2

Note: Award A1 for a parabola opening upward, A1 for vertex and intercepts in approximately correct positions.

[2 marks]

c.

$$\left( {\begin{array}{*{20}{c}} p\\ q \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { – 1}\\ { – 12} \end{array}} \right)$$
, $$t = 3$$ (accept $$p = – 1$$ , $$q = – 12$$ , $$t = 3$$ )Â Â Â Â  A1A1A1Â Â Â Â  N3

[3 marks]

d.

## Question

The following diagram shows part of the graph of f , where $$f(x) = {x^2} – x – 2$$ .

Find both x-intercepts.

[4]
a.

Find the x-coordinate of the vertex.

[2]
b.

## Markscheme

evidence of attempting to solve $$f(x) = 0$$Â Â Â  Â (M1)

evidence of correct workingÂ Â Â Â  A1

e.g. $$(x + 1)(x – 2)$$ , $$\frac{{1 \pm \sqrt 9 }}{2}$$

intercepts are $$( – 1{\text{, }}0)$$ and $$(2{\text{, }}0)$$ (accept $$x = – 1$$ , $$x = 2$$ ) A1A1Â Â Â Â  N1N1

[4 marks]

a.

evidence of appropriate methodÂ Â Â Â  (M1)

e.g. $${x_v} = \frac{{{x_1} + {x_2}}}{2}$$ , $${x_v} = – \frac{b}{{2a}}$$ ,Â reference to symmetry

$${x_v} = 0.5$$Â Â  Â A1 Â  Â  N2

[2 marks]

b.

## Question

Let $$f(x) = 8x – 2{x^2}$$ . Part of the graph of f is shown below.

Find the x-intercepts of the graph.

[4]
a.

(i) Â  Â  Write down the equation of the axis of symmetry.

(ii)Â Â Â  Find the y-coordinate of the vertex.

[3]
b(i) and (ii).

## Markscheme

evidence of setting function to zeroÂ Â Â Â  (M1)

e.g. $$f(x) = 0$$ , Â $$8x = 2{x^2}$$

evidence of correct workingÂ Â Â Â  A1

e.g. $$0 = 2x(4 – x)$$ , $$\frac{{ – 8 \pm \sqrt {64} }}{{ – 4}}$$

x-intercepts are at 4 and 0 (accept (4, 0)Â and (0, 0)Â , or $$x = 4$$Â , $$x = 0$$Â )Â Â Â Â  A1A1Â Â Â Â  N1N1

[4 marks]

a.

(i) $$x = 2$$Â (must be equation)Â Â Â Â  A1 Â  Â  N1

(ii) substituting $$x = 2$$Â into $$f(x)$$Â Â Â Â Â (M1)

$$y = 8$$Â Â Â Â  A1Â Â Â Â  N2

[3 marks]

b(i) and (ii).

## Question

Let $$f(x) = p(x – q)(x – r)$$ . Part of the graph of f is shown below.

The graph passes through the points (âˆ’2, 0), (0, âˆ’ 4) and (4, 0) .

Write down the value of q and of r.

[2]
a.

Write down the equation of the axis of symmetry.

[1]
b.

Find the value of p.

[3]
c.

## Markscheme

$$q = – 2$$ , $$r = 4$$ or $$q = 4$$ , $$r = – 2$$Â Â Â Â  A1A1Â Â Â Â  N2

[2 marks]

a.

$$x = 1$$ (must be an equation)Â Â Â Â  A1Â Â Â Â  N1

[1 mark]

b.

substituting $$(0{\text{, }} –Â 4)$$ into the equationÂ Â Â Â  (M1)

e.g. $$– 4 = p(0 – ( – 2))(0 – 4)$$ , $$– 4 = p( – 4)(2)$$

correct working towards solutionÂ Â Â Â  (A1)

e.g. $$– 4 = – 8p$$

$$p = \frac{4}{8}$$Â $$\left( { = \frac{1}{2}} \right)$$Â Â Â Â Â A1Â Â Â Â  N2

[3 marks]

c.

## Question

LetÂ $$f(x) = {x^2} + 4$$ and $$g(x) = x – 1$$Â .

FindÂ $$(f \circ g)(x)$$ .

[2]
a.

The vector $$\left( {\begin{array}{*{20}{c}} 3\\ { – 1} \end{array}} \right)$$ translates the graph of $$(f \circ g)$$ to the graph of h .

Find the coordinates of the vertex of the graph of h .

[3]
b.

The vector $$\left( {\begin{array}{*{20}{c}} 3\\ { – 1} \end{array}} \right)$$ translates the graph of $$(f \circ g)$$ to the graph of h .

Show that $$h(x) = {x^2} – 8x + 19$$ .

[2]
c.

The vector $$\left( {\begin{array}{*{20}{c}} 3\\ { – 1} \end{array}} \right)$$ translates the graph of $$(f \circ g)$$ to the graph of h .

The line $$y = 2x – 6$$ is a tangent to the graph of h at the point P. Find the x-coordinate of P.

[5]
d.

## Markscheme

attempt to form composition (in any order)Â Â Â Â  (M1)

$$(f \circ g)(x) = {(x – 1)^2} + 4$$Â Â Â  $$({x^2} – 2x + 5)$$Â Â Â Â Â A1Â Â Â Â  N2

[2 marks]

a.

METHOD 1

vertex of $$f \circ g$$ at (1, 4)Â Â Â Â  (A1)

evidence of appropriate approachÂ Â Â Â  (M1)

e.g. adding $$\left( {\begin{array}{*{20}{c}} 3\\ { – 1} \end{array}} \right)$$ to the coordinates of the vertex of $$f \circ g$$

vertex of h at (4, 3)Â Â Â Â  A1 Â  Â  N3

METHOD 2

attempt to find $$h(x)$$Â Â Â  Â (M1)

e.g. $${((x – 3) – 1)^2} + 4 – 1$$ , $$h(x) = (f \circ g)(x – 3) – 1$$

$$h(x) = {(x – 4)^2} + 3$$Â Â Â Â Â (A1)

vertex of h at (4, 3)Â Â Â Â  A1 Â  Â  N3

[3 marks]

b.

evidence of appropriate approachÂ Â Â Â  (M1)

e.g. $${(x – 4)^2} + 3$$ ,$${(x – 3)^2} – 2(x – 3) + 5 – 1$$

simplifyingÂ Â Â Â  A1

e.g. $$h(x) = {x^2} – 8x + 16 + 3$$ , $${x^2} – 6x + 9 – 2x + 6 + 4$$

$$h(x) = {x^2} – 8x + 19$$Â Â Â Â Â AG Â  Â  N0

[2 marks]

c.

METHOD 1

equating functions to find intersection pointÂ Â Â Â  (M1)

e.g. $${x^2} – 8x + 19 = 2x – 6$$ , $$y = h(x)$$

$${x^2} – 10x + 25 + 0$$Â Â Â Â  A1

evidence of appropriate approach to solveÂ Â Â Â  (M1)

appropriate workingÂ Â Â Â  A1

e.g. $${(x – 5)^2} = 0$$

$$x = 5$$Â  $$(p = 5)$$Â Â Â Â Â A1Â Â Â Â  N3

METHOD 2

attempt to find $$h'(x)$$Â Â Â Â Â (M1)

$$h(x) = 2x – 8$$Â Â Â Â  A1

recognizing that the gradient of the tangent is the derivativeÂ Â Â Â  (M1)

e.g. gradient at $$p = 2$$

$$2x – 8 = 2$$Â  $$(2x = 10)$$Â Â Â Â Â A1

$$x = 5$$Â Â Â Â Â A1 Â  Â  N3

[5 marks]

d.

## Examiners report

Candidates showed good understanding of finding the composite function in part (a).

a.

There were some who did not seem to understand what the vector translation meant in part (b).

b.

Candidates showed good understanding of manipulating the quadratic in part (c).

c.

There was more than one method to solve for h in part (d), and a pleasing number of candidates were successful in this part of the question.

d.

## Question

The following diagram shows part of the graph of a quadratic function f .

The x-intercepts are at $$( – 4{\text{, }}0)$$ and $$(6{\text{, }}0)$$ , and the y-intercept is at $$(0{\text{, }}240)$$ .

Write down $$f(x)$$ in the form $$f(x) = – 10(x – p)(x – q)$$ .

[2]
a.

Find another expression for $$f(x)$$ in the form $$f(x) = – 10{(x – h)^2} + k$$ .

[4]
b.

Show that $$f(x)$$ can also be written in the form $$f(x) = 240 + 20x – 10{x^2}$$ .

[2]
c.

A particle moves along a straight line so that its velocity, $$v{\text{ m}}{{\text{s}}^{ – 1}}$$ , at time t seconds is given by $$v = 240 + 20t – 10{t^2}$$ , for $$0 \le t \le 6$$ .

(i) Â  Â  Find the value of t when the speed of the particle is greatest.

(ii) Â Â  Find the acceleration of the particle when its speed is zero.

[7]
d(i) and (ii).

## Markscheme

$$f(x) = – 10(x + 4)(x – 6)$$Â Â Â Â  A1A1 Â  Â  N2

[2 marks]

a.

METHOD 1

attempting to find the x-coordinate of maximum pointÂ Â Â Â  (M1)

e.g. averaging the x-intercepts, sketch, $$y’ = 0$$ , axis of symmetry

attempting to find the y-coordinate of maximum pointÂ Â Â Â  (M1)

e.g. $$k = – 10(1 + 4)(1 – 6)$$

$$f(x) = – 10{(x – 1)^2} + 250$$Â Â Â Â Â A1A1 Â  Â  N4

METHOD 2

attempt to expand $$f(x)$$Â Â Â Â Â (M1)

e.g. $$– 10({x^2} – 2x – 24)$$

attempt to complete the squareÂ Â Â Â  (M1)

e.g. $$– 10({(x – 1)^2} – 1 – 24)$$

$$f(x) = – 10{(x – 1)^2} + 250$$Â Â Â  Â A1A1Â Â Â Â  N4

[4 marks]

b.

attempt to simplifyÂ Â Â Â  (M1)

e.g. distributive property, $$– 10(x – 1)(x – 1) + 250$$

correct simplificationÂ Â Â Â  A1

e.g. $$– 10({x^2} – 6x + 4x – 24)$$ , $$– 10({x^2} – 2x + 1) + 250$$

$$f(x) = 240 + 20x – 10{x^2}$$Â Â Â Â Â AGÂ Â Â Â  N0

[2 marks]

c.

(i) valid approachÂ Â Â Â  (M1)

e.g. vertex of parabola, $$v'(t) = 0$$

$$t = 1$$Â Â Â Â Â A1Â Â Â Â  N2

(ii) recognizing $$a(t) = v'(t)$$Â Â Â Â Â (M1)

$$a(t) = 20 – 20t$$Â Â Â Â Â A1A1

speed is zero $$\Rightarrow t = 6$$Â Â Â Â Â (A1)

$$a(6) = – 100$$ ($${\text{m}}{{\text{s}}^{ – 2}}$$)Â Â Â Â  A1Â Â Â Â  N3

[7 marks]

d(i) and (ii).

## Question

Let $$f$$ be a quadratic function. Part of the graph of $$f$$ is shown below.

The vertex is at P($$4$$, $$2$$) and the y-intercept is at Q($$0$$, $$6$$) .

Write down the equation of the axis of symmetry.

[1]
a.

The function f can be written in the form $$f(x) = a{(x – h)^2} + k$$ .

Write down the value of h and of k .

[2]
b.

The function f can be written in the form $$f(x) = a{(x – h)^2} + k$$ .

Find a .

[3]
c.

## Markscheme

$$x = 4$$ (must be an equation)Â Â Â Â  A1Â Â Â Â  N1

[1 mark]

a.

$$h = 4$$ , $$k = 2$$Â Â Â Â  A1A1Â Â Â Â  N2

[2 marks]

b.

attempt to substitute coordinates of any point on the graph into fÂ Â Â Â  (M1)

e.g. $$f(0) = 6$$ , $$6 = a{(0 – 4)^2} + 2$$ , $$f(4) = 2$$

correct equation (do not accept an equation that results from $$f(4) = 2$$ )Â Â Â Â  (A1)

e.g. $$6 = a{( – 4)^2} + 2$$ , $$6 = 16a + 2$$

$$a = \frac{4}{{16}}\left( { = \frac{1}{4}} \right)$$Â Â Â Â  A1 Â  Â  N2

[3 marks]

c.

## Examiners report

A surprising number of candidates missed part (a) of this question, which required them to write the equation of the axis of symmetry. Some candidates did not write their answer as an equation, while others simply wrote the formula $$x = – \frac{b}{{2a}}$$Â .

a.

This was answered correctly by the large majority of candidates.

b.

The rest of this question was answered correctly by the large majority of candidates. The mistakes seen in part (c) were generally due to either incorrect substitution of a point into the equation, or substitution of the vertex coordinates, which got the candidates nowhere.

c.

## Question

The following diagram shows the graph of a quadratic function f , for $$0 \le x \le 4$$ .

The graph passes through the point P(0, 13) , and its vertex is the point V(2, 1) .

The function can be written in the form $$f(x) = a{(x – h)^2} + k$$ .

(i)Â Â Â Â  Write down the value of h and of k .

(ii)Â Â Â  Show that $$a = 3$$ .

[4]
a(i) and (ii).

Find $$f(x)$$ Â , giving your answer in the form $$A{x^2} + Bx + C$$ .

[3]
b.

Calculate the area enclosed by the graph of f , the x-axis, and the lines $$x = 2$$ and $$x = 4$$ .

[8]
c.

## Markscheme

(i)Â $$h = 2$$ , $$k = 1$$Â Â Â  Â A1A1Â Â Â Â  N2

(ii) attempt to substitute coordinates of any point (except the vertex) on the graph into fÂ Â Â Â  M1

e.g. $$13 = a{(0 – 2)^2} + 1$$

working towards solutionÂ Â Â Â  A1

e.g. $$13 = 4a + 1$$

$$a = 3$$Â Â Â Â  AGÂ Â Â Â  N0

[4 marks]

a(i) and (ii).

attempting to expand their binomialÂ Â Â Â  (M1)

e.g. $$f(x) = 3({x^2} – 2 \times 2x + 4) + 1$$ , $${(x – 2)^2} = {x^2} – 4x + 4$$

correct workingÂ Â Â Â  (A1)

e.g. $$f(x) = 3{x^2} – 12x + 12 + 1$$

$$f(x) = 3{x^2} – 12x + 13$$Â (accept $$A = 3$$ , $$B = – 12$$ , $$C = 13$$Â )Â Â Â Â  A1Â Â Â Â  N2

[3 marks]

b.

METHOD 1

integral expressionÂ Â Â Â  (A1)

e.g. $$\int_2^4 {(3{x^2}}Â – 12x + 13)$$ , $$\int {f{\rm{d}}x}$$

$${\rm{Area}} = [{x^3} – 6{x^2} + 13x]_2^4$$Â Â Â  Â A1A1A1

Note: Award A1 for $${x^3}$$Â , A1 for $$– 6{x^2}$$Â , A1 for $$13x$$ .

correct substitution of correct limits into their expressionÂ Â Â Â  A1A1

e.g. $$({4^3} – 6 \times {4^2} + 13 \times 4) – ({2^3} – 6 \times {2^2} + 13 \times 2)$$ , $$64 – 96 + 52 – (8 – 24 + 26)$$

Note: Award A1 for substituting 4, A1 for substituting 2.

correct workingÂ Â Â Â  (A1)

e.g. $$64 – 96 + 52 – 8 + 24 – 26,20 – 10$$

$${\rm{Area}} = 10$$Â Â Â  Â A1Â Â Â Â  N3

[8 marks]

METHOD 2

integral expressionÂ Â Â Â  (A1)

e.g. $$\int_2^4 {(3{{(x – 2)}^2}}Â + 1)$$ , $$\int {f{\rm{d}}x}$$

$${\rm{Area}} = [{(x – 2)^3} + x]_2^4$$Â Â Â Â  A2A1

Note: Award A2 for $${(x – 2)^3}$$Â , A1 for $$x$$Â .

correct substitution of correct limits into their expressionÂ Â Â Â  A1A1

e.g. $${(4 – 2)^3} + 4 – [{(2 – 2)^3} + 2]$$ , $${2^3} + 4 – ({0^3} + 2)$$ , $${2^3} + 4 – 2$$

Note: Award A1 for substituting 4, A1 for substituting 2.

Â

correct workingÂ Â Â Â  (A1)

e.g. $$8 + 4 – 2$$

$${\rm{Area}} = 10$$Â  Â  A1Â Â Â Â  N3

[8 marks]

METHOD 3

recognizing area from 0 to 2 is same as area from 2 to 4Â Â Â Â  (R1)

e.g. sketch, $$\int_2^4 {f = \int_0^2 f }$$

integral expressionÂ Â Â Â  (A1)

e.g. $$\int_0^2 {(3{x^2}}Â – 12x + 13)$$ , $$\int {f{\rm{d}}x}$$

$${\rm{Area}} = [{x^3} – 6{x^2} + 13x]_0^2$$Â Â  Â  A1A1A1

Note: Award A1 for $${x^3}$$Â , A1 for $$– 6{x^2}$$Â , A1 for $$13x$$ .

correct substitution of correct limits into their expressionÂ Â Â Â  A1(A1)

e.g. $$({2^3} – 6 \times {2^2} + 13 \times 2) – ({0^3} – 6 \times {0^2} + 13 \times 0)$$ , $$8 – 24 + 26$$

Note: Award A1 for substituting 2, (A1) for substituting 0.

$${\rm{Area}} = 10$$Â Â Â  Â A1Â Â Â Â  N3

[8 marks]

c.

## Examiners report

In part (a), nearly all the candidates recognized that h and k were the coordinates of the vertex of the parabola, and most were able to successfully show that $$a = 3$$Â . Unfortunately, aÂ few candidates did not understand the “show that” command, and simply verified thatÂ $$a = 3$$ would work, rather than showing how to findÂ $$a = 3$$ .

a(i) and (ii).

In part (b), most candidates were able to find $$f(x)$$Â in the required form. For a few candidates, algebraic errors kept them from finding the correct function, even though they started with correct values for a, h and k.

b.

In part (c), nearly all candidates knew that they needed to integrate to find the area, but errors in integration, and algebraic and arithmetic errors prevented many from finding the correct area.

c.

## Question

The diagram below shows part of the graph of $$f(x) = (x – 1)(x + 3)$$ .

(a)Â Â Â Â  Write down the $$x$$-intercepts of the graph of $$f$$ .

(b)Â Â Â Â  Find the coordinates of the vertex of the graph of $$f$$ .

[6]
.

Write down the $$x$$-intercepts of the graph of $$f$$ .

[2]
a.

Find the coordinates of the vertex of the graph of $$f$$ .

[4]
b.

## Markscheme

(a)Â Â Â Â  $$x = 1$$ , $$x = – 3$$ (accept ($$1$$, $$0$$), ($$– 3$$, $$0$$) )Â Â Â  Â A1A1Â Â Â Â  N2

[2 marks]

Â

(b) Â  Â  METHOD 1

attempt to find $$x$$-coordinateÂ Â Â Â  (M1)

egÂ Â  $$\frac{{1 + – 3}}{2}$$Â , $$x = \frac{{ – b}}{{2a}}$$Â , $$f'(x) = 0$$

correct value, $$x = – 1$$Â (may be seen as a coordinate in the answer)Â Â Â Â  A1

attempt to find their $$y$$-coordinateÂ Â Â Â  (M1)

egÂ Â  $$f( – 1)$$ , $$– 2 \times 2$$ , $$y = \frac{{ – D}}{{4a}}$$

$$y = – 4$$Â Â Â Â  A1

vertex ($$– 1$$, $$– 4$$)Â Â Â  Â N3Â Â

METHOD 2

attempt to complete the squareÂ Â Â Â  (M1)

egÂ Â  $${x^2} + 2x + 1 – 1 – 3$$Â

attempt to put into vertex formÂ Â Â Â  (M1)

egÂ Â  $${(x + 1)^2} – 4$$ , $${(x – 1)^2} + 4$$

vertex ($$– 1$$, $$– 4$$)Â Â Â  Â A1A1Â Â Â Â  N3

[4 marks]

.

$$x = 1$$ , $$x = – 3$$ (accept ($$1$$, $$0$$), ($$– 3$$, $$0$$) )Â Â Â  Â A1A1Â Â Â Â  N2

[2 marks]

Â

a.

METHOD 1

attempt to find $$x$$-coordinateÂ Â Â Â  (M1)

egÂ Â  $$\frac{{1 + – 3}}{2}$$Â , $$x = \frac{{ – b}}{{2a}}$$Â , $$f'(x) = 0$$

correct value, $$x = – 1$$Â (may be seen as a coordinate in the answer)Â Â Â Â  A1

attempt to find their $$y$$-coordinateÂ Â Â Â  (M1)

egÂ Â  $$f( – 1)$$ , $$– 2 \times 2$$ , $$y = \frac{{ – D}}{{4a}}$$

$$y = – 4$$Â Â Â Â  A1

vertex ($$– 1$$, $$– 4$$)Â Â Â  Â N3Â Â

METHOD 2

attempt to complete the squareÂ Â Â Â  (M1)

egÂ Â  $${x^2} + 2x + 1 – 1 – 3$$Â

attempt to put into vertex formÂ Â Â Â  (M1)

egÂ Â  $${(x + 1)^2} – 4$$ , $${(x – 1)^2} + 4$$

vertex ($$– 1$$, $$– 4$$)Â Â Â  Â A1A1Â Â Â Â  N3

[4 marks]

b.

## Question

LetÂ $$f(x) = \sin x + \frac{1}{2}{x^2} – 2x$$ , forÂ $$0 \le x \le \pi$$ .

Let $$g$$ be a quadratic function such that $$g(0) = 5$$ . The line $$x = 2$$ is the axis of symmetry of the graph of $$g$$ .

The function $$g$$ can be expressed in the form $$g(x) = a{(x – h)^2} + 3$$ .

Find $$f'(x)$$ .

[3]
a.

FindÂ $$g(4)$$ .

[3]
b.

(i)Â Â Â Â  Write down the value of $$h$$ .

(ii)Â Â Â Â  Find the value of $$a$$ .

[4]
c.

Find the value of $$x$$ for which the tangent to the graph of $$f$$ is parallel to the tangent to the graph of $$g$$ .

[6]
d.

## Markscheme

$$f'(x) = \cos x + x – 2$$Â Â Â Â  A1A1A1Â Â Â Â  N3

Note: Award A1 for each term.

[3 marks]

a.

recognizing $$g(0) = 5$$Â gives the point ($$0$$, $$5$$)Â Â Â  Â (R1)

recognize symmetryÂ Â Â Â  (M1)

eg vertex, sketch

$$g(4) = 5$$Â Â Â Â  A1Â Â Â Â  N3

[3 marks]

b.

(i)Â Â Â Â  $$h = 2$$Â Â Â Â  A1 N1

(ii)Â Â Â Â  substituting intoÂ $$g(x) = a{(x – 2)^2} + 3$$Â (not the vertex)Â Â Â Â  (M1)

eg Â  $$5 = a{(0 – 2)^2} + 3$$ , $$5 = a{(4 – 2)^2} + 3$$

working towards solutionÂ Â Â Â  (A1)

eg Â  $$5 = 4a + 3$$ , $$4a = 2$$

$$a = \frac{1}{2}$$Â Â Â Â  A1Â Â Â Â  N2

[4 marks]

c.

$$g(x) = \frac{1}{2}{(x – 2)^2} + 3 = \frac{1}{2}{x^2} – 2x + 5$$

correct derivative of $$g$$Â Â Â Â  A1A1

eg Â  $$2 \times \frac{1}{2}(x – 2)$$ , $$x – 2$$

evidence of equating both derivativesÂ Â Â Â  (M1)

eg Â  $$f’ = g’$$

correct equationÂ Â Â Â  (A1)

eg Â  $$\cos x + x – 2 = x – 2$$

working towards a solutionÂ Â Â Â  (A1)

egÂ Â  $$\cos x = 0$$Â , combining like terms

$$x = \frac{\pi }{2}$$Â Â Â  A1Â Â Â Â  N0

Note: Do not award final A1 if additional values are given.

[6 marks]

d.

## Question

Let $$f(x) = a{(x – h)^2} + k$$. The vertex of the graph of $$f$$ is at $$(2, 3)$$ and the graph passesÂ through $$(1, 7)$$.

Write down the value ofÂ $$h$$Â and ofÂ $$k$$.

[2]
a.

Find the value of $$a$$.

[3]
b.

## Markscheme

$$h = 2,{\text{ }}k = 3$$ Â  Â Â A1A1 Â  Â  N2

[2 marks]

a.

attempt to substitute $$(1,7)$$ in any order into theirÂ $$f(x)$$Â  Â  Â (M1)

eg Â  Â Â $$7 = a{(1 – 2)^2} + 3{\text{, }}7 = a{(1 – 3)^2} + 2{\text{, }}1 = a{(7 – 2)^2} + 3$$

correct equation Â  Â  (A1)

eg Â  Â Â $$7 = a + 3$$

a = 4 Â  Â Â A1 Â  Â  N2

[3 marks]

b.

## Question

Let $$f(x) = 3{x^2} – 6x + p$$. The equation $$f(x) = 0$$ has two equal roots.

Write down the value of the discriminant.

[2]
a(i).

Hence, show that $$p = 3$$.

[1]
a(ii).

The graph of $$f$$has its vertex on the $$x$$-axis.

Find the coordinates of the vertex of the graph of $$f$$.

[4]
b.

The graph of $$f$$ has its vertex on the $$x$$-axis.

Write down the solution of $$f(x) = 0$$.

[1]
c.

The graph of $$f$$Â has its vertex on the $$x$$-axis.

The function can be written in the form $$f(x) = a{(x – h)^2} + k$$. Write down the value of $$a$$.

[1]
d(i).

The graph of $$f$$ has its vertex on the $$x$$-axis.

The function can be written in the form $$f(x) = a{(x – h)^2} + k$$. Write down the value of $$h$$.

[1]
d(ii).

The graph of $$f$$ has its vertex on the $$x$$-axis.

The function can be written in the form $$f(x) = a{(x – h)^2} + k$$. Write down the value of $$k$$.

[1]
d(iii).

The graph of $$f$$ has its vertex on the $$x$$-axis.

The graph of a function $$g$$ is obtained from the graph of $$f$$ by a reflection of $$f$$ in the $$x$$-axis, followed by a translation by the vector $$\left( \begin{array}{c}0\\6\end{array} \right)$$. Find $$g$$, giving your answer in the form $$g(x) = A{x^2} + Bx + C$$.

[4]
e.

## Markscheme

correct value $$0$$, or $$36 – 12p$$ Â  Â  A2 Â  Â  N2

[2 marks]

a(i).

correct equation which clearly leads to $$p = 3$$ Â  Â  A1

eg Â  Â  $$36 – 12p = 0,{\text{ }}36 = 12p$$

$$p = 3$$ Â  Â  AG Â  Â  N0

[1 mark]

a(ii).

METHOD 1

valid approach Â  Â  (M1)

eg Â  Â  $$x =Â – \frac{b}{{2a}}$$

correct working Â  Â  A1

eg Â  Â  $$– \frac{{( – 6)}}{{2(3)}},{\text{ }}x = \frac{6}{6}$$

correct answers Â  Â  A1A1 Â  Â  N2

eg Â  Â  $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$

METHOD 2

valid approach Â  Â  (M1)

eg Â  Â  $$f(x) = 0$$, factorisation, completing the square

correct working Â  Â  A1

eg Â  Â  $${x^2} – 2x + 1 = 0,{\text{ }}(3x – 3)(x – 1),{\text{ }}f(x) = 3{(x – 1)^2}$$

correct answers Â  Â  A1A1 Â  Â  N2

eg Â  Â  $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$

METHOD 3

valid approach using derivative Â  Â  (M1)

eg Â  Â  $$f'(x) = 0,{\text{ }}6x – 6$$

correct equation Â  Â  A1

eg Â  Â  $$6x – 6 = 0$$

correct answers Â  Â  A1A1 Â  Â  N2

eg Â  Â  $$x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)$$

[4 marks]

b.

$$x = 1$$ Â  Â  A1 Â  Â  N1

[1 mark]

c.

$$a = 3$$ Â  Â  A1 Â  Â  N1

[1 mark]

d(i).

$$h = 1$$ Â  Â  A1 Â  Â  N1

[1 mark]

d(ii).

$$k = 0$$ Â  Â  A1 Â  Â  N1

[1 mark]

d(iii).

attempt to apply vertical reflection Â  Â  (M1)

eg Â  Â Â $$– f(x),{\text{ }} – 3{(x – 1)^2}$$, sketch

attempt to apply vertical shift 6 units up Â  Â  (M1)

eg Â  Â Â $$– f(x) + 6$$, vertex $$(1, 6)$$

transformations performed correctly (in correct order) Â  Â  (A1)

eg Â  Â Â $$– 3{(x – 1)^2} + 6,{\text{ }} – 3{x^2} + 6x – 3 + 6$$

$$g(x) =Â – 3{x^2} + 6x + 3$$ Â  Â  A1 Â  Â  N3

[4 marks]

e.

## Question

Let $$f(x) = {x^2} + x – 6$$.

Write down the $$y$$-intercept of the graph of $$f$$.

[1]
a.

Solve $$f(x) = 0$$.

[3]
b.

On the following grid, sketch the graph of $$f$$, for $$– 4 \le x \le 3$$.

[3]
c.

## Markscheme

$$y$$-intercept is $$– 6,{\text{ }}(0,{\text{ }} – 6),{\text{ }}y =Â – 6$$ Â  Â  A1

[1 mark]

a.

valid attempt to solve Â  Â  (M1)

eg$$\;\;\;(x – 2)(x + 3) = 0,{\text{ }}x = \frac{{ – 1 \pm \sqrt {1 + 24} }}{2}$$, one correct answer

$$x = 2,{\text{ }}x =Â – 3$$ Â  Â  A1A1 Â  Â  N3

[3 marks]

b.

Â  Â Â A1A1A1

Note: Â  Â  The shape must be an approximately correct concave up parabola. Only if the shape is correct, award the following:

A1 for the $$y$$-intercept in circle and the vertex approximately on $$x =Â – \frac{1}{2}$$, below $$y =Â – 6$$,

A1 for both theÂ $$x$$-intercepts in circles,

A1 for both end points in ovals.

[3 marks]

Total [7 marks]

c.

## Question

Consider $$f(x) = {x^2} + qx + r$$. The graph of $$f$$ has a minimum value when $$x = Â – 1.5$$.

The distance between the two zeros of $$f$$ is 9.

Show that the two zeros are 3 and $$– 6$$.

[2]
a.

Find the value of $$q$$ and of $$r$$.

[4]
b.

## Markscheme

recognition that the $$x$$-coordinate of the vertex is $$– 1.5$$Â (seen anywhere) Â  Â  (M1)

eg$$\,\,\,\,\,$$axis of symmetry is $$– 1.5$$, sketch,Â $$f'( – 1.5) = 0$$

correct working to find the zeroes Â  Â  A1

eg$$\,\,\,\,\,$$$$– 1.5 \pm 4.5$$

$$x = Â – 6$$ and $$x = 3$$Â Â  Â  AG Â  Â  N0

[2 marks]

a.

METHOD 1 (using factors)

attempt to write factors Â  Â  (M1)

eg$$\,\,\,\,\,$$$$(x – 6)(x + 3)$$

correct factors Â  Â  A1

eg$$\,\,\,\,\,$$$$(x – 3)(x + 6)$$

$$q = 3,{\text{ }}r = Â – 18$$ Â  Â A1A1 Â  Â  N3

METHOD 2 (using derivative or vertex)

valid approach to find $$q$$ Â  Â  (M1)

eg$$\,\,\,\,\,$$$$f'( – 1.5) = 0,{\text{ }} – \frac{q}{{2a}} = Â – 1.5$$

$$q = 3$$ Â  Â A1

correct substitution Â  Â  A1

eg$$\,\,\,\,\,$$$${3^2} + 3(3) + r = 0,{\text{ }}{( – 6)^2} + 3( – 6) + r = 0$$

$$r = Â – 18$$ Â  Â A1

$$q = 3,{\text{ }}r = Â – 18$$ Â  Â N3

METHOD 3 (solving simultaneously)

valid approach setting up system of two equations Â  Â  (M1)

eg$$\,\,\,\,\,$$$$9 + 3q + r = 0,{\text{ }}36 – 6q + r = 0$$

one correct value

eg$$\,\,\,\,\,$$$$q = 3,{\text{ }}r = Â – 18$$Â Â  Â  A1

correct substitution Â  Â  A1

eg$$\,\,\,\,\,$$$${3^2} + 3(3) + r = 0,{\text{ }}{( – 6)^2} + 3( – 6) + r = 0,{\text{ }}{3^2} + 3q – 18 = 0,{\text{ }}36 – 6q – 18 = 0$$

second correct value Â  Â  A1

eg$$\,\,\,\,\,$$$$q = 3,{\text{ }}r = Â – 18$$

$$q = 3,{\text{ }}r = Â – 18$$ Â  Â N3

[4 marks]

b.

## Examiners report

As a â€˜show thatâ€™ question, part a) required a candidate to independently find the answers. Again, too many candidates used the given answers (of 3 and $$– 6$$) to show that the two zeros were 3 and $$– 6$$ (a circular argument). Those who were able to recognize that the $$x$$-coordinate of the vertex is $$– 1.5$$ tended to then use the given answers and work backwards thus scoring no further marks in part a).

a.

Answers to part b) were more successful with a good variety of methods used and correct solutions seen.

b.

## Question

Let f(x) = ax2 âˆ’ 4x âˆ’ c. A horizontal line, L , intersects the graph of f at x = âˆ’1 and x = 3.

The equation of the axis of symmetry is x = p. Find p.

[2]
a.i.

Hence, show that a = 2.

[2]
a.ii.

The equation of L is y = 5 . Find the value of c.

[3]
b.

## Markscheme

METHOD 1 (using symmetry to findÂ p)

valid approachÂ  Â  Â  (M1)

egÂ Â $$\frac{{ – 1 + 3}}{2}$$,Â

p = 1Â  Â  Â A1 N2

Note: Award no marks if they work backwards by substituting a = 2 intoÂ $$– \frac{b}{{2a}}$$ to find p.

Do not acceptÂ $$p = \frac{2}{a}$$.

METHOD 2 (calculating a first)
(i) & (ii) valid approach to calculate aÂ  Â  Â  M1

egÂ  Â a + 4Â âˆ’ c = a(32) âˆ’ 4(3)Â âˆ’ c,Â Â f(âˆ’1) = f(3)

correct workingÂ  Â  Â  A1

egÂ  Â 8a = 16

a = 2Â  Â  Â  AG N0

valid approach to find pÂ  Â  Â Â (M1)

egÂ  Â $$– \frac{b}{{2a}},\,\,\,\frac{4}{{2\left( 2 \right)}}$$

p = 1Â  Â  Â  A1 N2

[2 marks]

a.i.

METHOD 1

valid approachÂ  Â  Â  Â M1

egÂ  $$– \frac{b}{{2a}},\,\,\,\frac{4}{{2a}}$$Â (might be seen in (i)), f’â€‰(1) = 0

correct equationÂ  Â  Â A1

egÂ  $$\frac{4}{{2a}}$$ = 1, 2a(1)Â âˆ’ 4 = 0

a = 2Â  Â  Â  AG N0

METHOD 2 (calculating a first)
(i) & (ii) valid approach to calculate aÂ  Â  Â  M1

egÂ  Â a + 4Â âˆ’ c = a(32) âˆ’ 4(3)Â âˆ’ c,Â Â f(âˆ’1) = f(3)

correct workingÂ  Â  Â  A1

egÂ  Â 8a = 16

a = 2Â  Â  Â  AG N0

[2 marks]

a.ii.

valid approachÂ  Â  Â  (M1)
egÂ  Â f(âˆ’1) = 5, f(3) =5

correct workingÂ  Â  Â  Â (A1)
egÂ  Â 2 + 4Â âˆ’ c = 5, 18Â âˆ’ 12Â âˆ’ c = 5

cÂ = 1Â  Â  Â A1 N2

[3 marks]

b.

## Question

Let $$f\left( x \right) = p{x^2} + qx – 4p$$, where pÂ â‰  0. FindÂ Find the number of roots for the equation $$f\left( x \right) = 0$$.

## Markscheme

METHOD 1

evidence of discriminantÂ  Â  Â  (M1)
egÂ Â $${b^2} – 4ac,\,\,\Delta$$

correct substitution into discriminantÂ  Â  Â  (A1)
egÂ Â $${q^2} – 4p\left( { – 4p} \right)$$

correct discriminantÂ  Â  Â  Â A1
egÂ Â $${q^2}Â + 16{p^2}$$

$$16{p^2} > 0\,\,\,\,\left( {{\text{accept}}\,\,{p^2} > 0} \right)$$Â  Â  Â A1

$${q^2} \geqslant 0\,\,\,\,\left( {{\text{do not accept}}\,\,{q^2} > 0} \right)$$Â  Â  Â A1

$${q^2} + 16{p^2} > 0$$Â  Â  Â  A1

$$f$$ has 2 rootsÂ  Â  Â A1 N0

METHOD 2

y-intercept = âˆ’4p (seen anywhere)Â  Â  Â  A1

if p is positive, then the y-intercept will be negativeÂ  Â  Â  A1

an upward-opening parabola with a negative y-interceptÂ  Â  Â  R1
egÂ  sketch that must indicate p > 0.

if p is negative, then the y-intercept will be positiveÂ  Â  Â  A1

a downward-opening parabola with a positive y-interceptÂ  Â  Â  R1
egÂ  sketch that must indicate pÂ >Â 0.

$$f$$ has 2 rootsÂ  Â  Â A2 N0

[7 marks]

## Question

LetÂ $$f(x) = 3{(x + 1)^2} – 12$$ .

Show that $$f(x) = 3{x^2} + 6x – 9$$ .

[2]
a.

For the graph of f

(i)Â Â Â Â  write down the coordinates of the vertex;

(ii)Â Â Â  write down the y-intercept;

(iii)Â Â  find both x-intercepts.

[7]
b(i), (ii) and (iii).

Hence sketch the graph of f .

[3]
c.

Let $$g(x) = {x^2}$$ . The graph of f may be obtained from the graph of g by the following two transformations

a stretch of scale factor t in the y-direction,

followed by a translation of $$\left( \begin{array}{l} p\\ q \end{array} \right)$$ .

Write down $$\left( \begin{array}{l} p\\ q \end{array} \right)$$ and the value of t .

[3]
d.

## Markscheme

$$f(x) = 3({x^2} + 2x + 1) – 12$$Â Â Â Â Â A1

$$= 3{x^2} + 6x + 3 – 12$$Â Â Â Â Â A1

$$= 3{x^2} + 6x – 9$$Â Â Â Â Â AGÂ Â Â Â  N0

[2 marks]

a.

(i) vertex is $$( – 1, – 12)$$Â Â Â Â Â A1A1Â Â Â Â  N2

(ii) $$y = – 9$$ , or $$(0, – 9)$$Â Â Â  Â A1 Â  Â  N1

(iii) evidence of solving $$f(x) = 0$$Â Â Â Â Â M1

e.g. factorizing, formula

correct workingÂ Â Â Â  A1

e.g. $$3(x + 3)(x – 1) = 0$$ , $$x = \frac{{ – 6 \pm \sqrt {36 + 108} }}{6}$$

$$x = – 3$$ , $$x = 1$$ , or $$( – 3{\text{, }}0){\text{, }}(1{\text{, }}0)$$Â Â Â Â Â A1A1 Â  Â  N2

[7 marks]

b(i), (ii) and (iii).

Â Â Â Â  A1A1A1 Â  Â  N3

Note: Award A1 for a parabola opening upward, A1 for vertex in approximately correct position, A1 for intercepts in approximately correct positions. Scale and labelling not required.

[3 marks]

c.

$$\left( \begin{array}{l} p\\ q \end{array} \right) = \left( \begin{array}{l} – 1\\ – 12 \end{array} \right)$$ , $$t = 3$$Â Â Â Â  A1A1A1Â Â Â Â  N3

[3 marks]

d.