# 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)     A1     N1

[1 mark]

a.

$$h = 4$$ , $$k = 2$$     A1A1     N2

[2 marks]

b.

attempt to substitute coordinates of any point on the graph into f     (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 $$f(4) = 2$$ )     (A1)

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

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

[3 marks]

c.

## Question

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

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.

Find $$g(4)$$ .

[3]
b.

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

(ii)     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

$$f'(x) = \cos x + x – 2$$     A1A1A1     N3

Note: Award A1 for each term.

[3 marks]

a.

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

recognize symmetry     (M1)

eg vertex, sketch

$$g(4) = 5$$     A1     N3

[3 marks]

b.

(i)     $$h = 2$$     A1 N1

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

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

working towards solution     (A1)

eg   $$5 = 4a + 3$$ , $$4a = 2$$

$$a = \frac{1}{2}$$     A1     N2

[4 marks]

c.

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

correct derivative of $$g$$     A1A1

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

evidence of equating both derivatives     (M1)

eg   $$f’ = g’$$

correct equation     (A1)

eg   $$\cos x + x – 2 = x – 2$$

working towards a solution     (A1)

eg   $$\cos x = 0$$ , combining like terms

$$x = \frac{\pi }{2}$$    A1     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 passes through $$(1, 7)$$.

Write down the value of $$h$$ and of $$k$$.

[2]
a.

Find the value of $$a$$.

[3]
b.

## Markscheme

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

[2 marks]

a.

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

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

correct equation     (A1)

eg     $$7 = a + 3$$

a = 4     A1     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 $$f$$ 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 $$36 – 12p$$     A2     N2

[2 marks]

a(i).

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

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

$$p = 3$$     AG     N0

[1 mark]

a(ii).

METHOD 1

valid approach     (M1)

eg     $$x = – \frac{b}{{2a}}$$

correct working     A1

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

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

METHOD 2

valid approach     (M1)

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

correct working     A1

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

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

METHOD 3

valid approach using derivative     (M1)

eg     $$f'(x) = 0,{\text{ }}6x – 6$$

correct equation     A1

eg     $$6x – 6 = 0$$

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

[4 marks]

b.

$$x = 1$$     A1     N1

[1 mark]

c.

$$a = 3$$     A1     N1

[1 mark]

d(i).

$$h = 1$$     A1     N1

[1 mark]

d(ii).

$$k = 0$$     A1     N1

[1 mark]

d(iii).

attempt to apply vertical reflection     (M1)

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

attempt to apply vertical shift 6 units up     (M1)

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

transformations performed correctly (in correct order)     (A1)

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

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

[4 marks]

e.

## Question

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

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 $$– 1.5$$ (seen anywhere)     (M1)

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

correct working to find the zeroes     A1

eg$$\,\,\,\,\,$$$$– 1.5 \pm 4.5$$

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

[2 marks]

a.

METHOD 1 (using factors)

attempt to write factors     (M1)

eg$$\,\,\,\,\,$$$$(x – 6)(x + 3)$$

correct factors     A1

eg$$\,\,\,\,\,$$$$(x – 3)(x + 6)$$

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

METHOD 2 (using derivative or vertex)

valid approach to find $$q$$     (M1)

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

$$q = 3$$    A1

correct substitution     A1

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

$$r = – 18$$    A1

$$q = 3,{\text{ }}r = – 18$$    N3

METHOD 3 (solving simultaneously)

valid approach setting up system of two equations     (M1)

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

one correct value

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

correct substitution     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     A1

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

$$q = 3,{\text{ }}r = – 18$$    N3

[4 marks]

b.

## Question

Let f(x) = ax2 − 4xc. A horizontal line, L , intersects the graph of f at x = −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 find p)

valid approach      (M1)

eg  $$\frac{{ – 1 + 3}}{2}$$,

p = 1     A1 N2

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

Do not accept $$p = \frac{2}{a}$$.

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

eg   a + 4 − c = a(32) − 4(3) − c,  f(−1) = f(3)

correct working      A1

eg   8a = 16

a = 2      AG N0

valid approach to find p      (M1)

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

p = 1      A1 N2

[2 marks]

a.i.

METHOD 1

valid approach       M1

eg  $$– \frac{b}{{2a}},\,\,\,\frac{4}{{2a}}$$ (might be seen in (i)), f’ (1) = 0

correct equation     A1

eg  $$\frac{4}{{2a}}$$ = 1, 2a(1) − 4 = 0

a = 2      AG N0

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

eg   a + 4 − c = a(32) − 4(3) − c,  f(−1) = f(3)

correct working      A1

eg   8a = 16

a = 2      AG N0

[2 marks]

a.ii.

valid approach      (M1)
eg   f(−1) = 5, f(3) =5

correct working       (A1)
eg   2 + 4 − c = 5, 18 − 12 − c = 5

c = 1     A1 N2

[3 marks]

b.

## Question

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

## Markscheme

METHOD 1

evidence of discriminant      (M1)
eg  $${b^2} – 4ac,\,\,\Delta$$

correct substitution into discriminant      (A1)
eg  $${q^2} – 4p\left( { – 4p} \right)$$

correct discriminant       A1
eg  $${q^2} + 16{p^2}$$

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

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

$${q^2} + 16{p^2} > 0$$      A1

$$f$$ has 2 roots     A1 N0

METHOD 2

y-intercept = −4p (seen anywhere)      A1

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

an upward-opening parabola with a negative y-intercept      R1
eg  sketch that must indicate p > 0.

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

a downward-opening parabola with a positive y-intercept      R1
eg  sketch that must indicate p > 0.

$$f$$ has 2 roots     A2 N0

[7 marks]

## 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)     write down the coordinates of the vertex;

(ii)    write down the y-intercept;

(iii)   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

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

$$= 3{x^2} + 6x + 3 – 12$$     A1

$$= 3{x^2} + 6x – 9$$     AG     N0

[2 marks]

a.

(i) vertex is $$( – 1, – 12)$$     A1A1     N2

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

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

e.g. factorizing, formula

correct working     A1

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

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

[7 marks]

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

A1A1A1     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.

$$\left( \begin{array}{l} p\\ q \end{array} \right) = \left( \begin{array}{l} – 1\\ – 12 \end{array} \right)$$ , $$t = 3$$     A1A1A1     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)     write down the coordinates of the vertex;

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

(iii)   write down the y-intercept;

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

[3]
d.

## Markscheme

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

$$= 3{x^2} + 6x + 3 – 12$$     A1

$$= 3{x^2} + 6x – 9$$     AG     N0

[2 marks]

a.

(i) vertex is $$( – 1{\text{, }} – 12)$$     A1A1     N2

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

(iii) $$(0{\text{, }} – 9)$$     A1     N1

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

e.g. factorizing, formula,

correct working     A1

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

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

[8 marks]

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

A1A1     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 $$p = – 1$$ , $$q = – 12$$ , $$t = 3$$ )     A1A1A1     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 $$f(x) = 0$$     (M1)

evidence of correct working     A1

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

intercepts are $$( – 1{\text{, }}0)$$ and $$(2{\text{, }}0)$$ (accept $$x = – 1$$ , $$x = 2$$ ) A1A1     N1N1

[4 marks]

a.

evidence of appropriate method     (M1)

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

$${x_v} = 0.5$$    A1     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)     Write down the equation of the axis of symmetry.

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

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

## Markscheme

evidence of setting function to zero     (M1)

e.g. $$f(x) = 0$$ ,  $$8x = 2{x^2}$$

evidence of correct working     A1

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

x-intercepts are at 4 and 0 (accept (4, 0) and (0, 0) , or $$x = 4$$ , $$x = 0$$ )     A1A1     N1N1

[4 marks]

a.

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

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

$$y = 8$$     A1     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 (−2, 0), (0, − 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 $$q = 4$$ , $$r = – 2$$     A1A1     N2

[2 marks]

a.

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

[1 mark]

b.

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

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

correct working towards solution     (A1)

e.g. $$– 4 = – 8p$$

$$p = \frac{4}{8}$$ $$\left( { = \frac{1}{2}} \right)$$     A1     N2

[3 marks]

c.

## Question

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

Find $$(f \circ g)(x)$$ .

[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)     (M1)

$$(f \circ g)(x) = {(x – 1)^2} + 4$$    $$({x^2} – 2x + 5)$$     A1     N2

[2 marks]

a.

METHOD 1

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

evidence of appropriate approach     (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)     A1     N3

METHOD 2

attempt to find $$h(x)$$     (M1)

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

$$h(x) = {(x – 4)^2} + 3$$     (A1)

vertex of h at (4, 3)     A1     N3

[3 marks]

b.

evidence of appropriate approach     (M1)

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

simplifying     A1

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

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

[2 marks]

c.

METHOD 1

equating functions to find intersection point     (M1)

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

$${x^2} – 10x + 25 + 0$$     A1

evidence of appropriate approach to solve     (M1)

appropriate working     A1

e.g. $${(x – 5)^2} = 0$$

$$x = 5$$  $$(p = 5)$$     A1     N3

METHOD 2

attempt to find $$h'(x)$$     (M1)

$$h(x) = 2x – 8$$     A1

recognizing that the gradient of the tangent is the derivative     (M1)

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

$$2x – 8 = 2$$  $$(2x = 10)$$     A1

$$x = 5$$     A1     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)     Find the value of t when the speed of the particle is greatest.

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

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

## Markscheme

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

[2 marks]

a.

METHOD 1

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

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

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

e.g. $$k = – 10(1 + 4)(1 – 6)$$

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

METHOD 2

attempt to expand $$f(x)$$     (M1)

e.g. $$– 10({x^2} – 2x – 24)$$

attempt to complete the square     (M1)

e.g. $$– 10({(x – 1)^2} – 1 – 24)$$

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

[4 marks]

b.

attempt to simplify     (M1)

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

correct simplification     A1

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

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

[2 marks]

c.

(i) valid approach     (M1)

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

$$t = 1$$     A1     N2

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

$$a(t) = 20 – 20t$$     A1A1

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

$$a(6) = – 100$$ ($${\text{m}}{{\text{s}}^{ – 2}}$$)     A1     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)     A1     N1

[1 mark]

a.

$$h = 4$$ , $$k = 2$$     A1A1     N2

[2 marks]

b.

attempt to substitute coordinates of any point on the graph into f     (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 $$f(4) = 2$$ )     (A1)

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

$$a = \frac{4}{{16}}\left( { = \frac{1}{4}} \right)$$     A1     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 $$x = – \frac{b}{{2a}}$$ .

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)     Write down the value of h and of k .

(ii)    Show that $$a = 3$$ .

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

Find $$f(x)$$  , 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) $$h = 2$$ , $$k = 1$$     A1A1     N2

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

e.g. $$13 = a{(0 – 2)^2} + 1$$

working towards solution     A1

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

$$a = 3$$     AG     N0

[4 marks]

a(i) and (ii).

attempting to expand their binomial     (M1)

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

correct working     (A1)

e.g. $$f(x) = 3{x^2} – 12x + 12 + 1$$

$$f(x) = 3{x^2} – 12x + 13$$ (accept $$A = 3$$ , $$B = – 12$$ , $$C = 13$$ )     A1     N2

[3 marks]

b.

METHOD 1

integral expression     (A1)

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

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

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

correct substitution of correct limits into their expression     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 working     (A1)

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

$${\rm{Area}} = 10$$     A1     N3

[8 marks]

METHOD 2

integral expression     (A1)

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

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

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

correct substitution of correct limits into their expression     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.

correct working     (A1)

e.g. $$8 + 4 – 2$$

$${\rm{Area}} = 10$$    A1     N3

[8 marks]

METHOD 3

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

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

integral expression     (A1)

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

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

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

correct substitution of correct limits into their expression     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.

$${\rm{Area}} = 10$$     A1     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 $$a = 3$$ . Unfortunately, a few candidates did not understand the “show that” command, and simply verified that $$a = 3$$ would work, rather than showing how to find $$a = 3$$ .

a(i) and (ii).

In part (b), most candidates were able to find $$f(x)$$ 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)     Write down the $$x$$-intercepts of the graph of $$f$$ .

(b)     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)     $$x = 1$$ , $$x = – 3$$ (accept ($$1$$, $$0$$), ($$– 3$$, $$0$$) )     A1A1     N2

[2 marks]

(b)     METHOD 1

attempt to find $$x$$-coordinate     (M1)

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

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

attempt to find their $$y$$-coordinate     (M1)

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

$$y = – 4$$     A1

vertex ($$– 1$$, $$– 4$$)     N3

METHOD 2

attempt to complete the square     (M1)

eg   $${x^2} + 2x + 1 – 1 – 3$$

attempt to put into vertex form     (M1)

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

vertex ($$– 1$$, $$– 4$$)     A1A1     N3

[4 marks]

.

$$x = 1$$ , $$x = – 3$$ (accept ($$1$$, $$0$$), ($$– 3$$, $$0$$) )     A1A1     N2

[2 marks]

a.

METHOD 1

attempt to find $$x$$-coordinate     (M1)

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

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

attempt to find their $$y$$-coordinate     (M1)

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

$$y = – 4$$     A1

vertex ($$– 1$$, $$– 4$$)     N3

METHOD 2

attempt to complete the square     (M1)

eg   $${x^2} + 2x + 1 – 1 – 3$$

attempt to put into vertex form     (M1)

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

vertex ($$– 1$$, $$– 4$$)     A1A1     N3

[4 marks]

b.

## Question

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

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.

Find $$g(4)$$ .

[3]
b.

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

(ii)     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

$$f'(x) = \cos x + x – 2$$     A1A1A1     N3

Note: Award A1 for each term.

[3 marks]

a.

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

recognize symmetry     (M1)

eg vertex, sketch

$$g(4) = 5$$     A1     N3

[3 marks]

b.

(i)     $$h = 2$$     A1 N1

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

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

working towards solution     (A1)

eg   $$5 = 4a + 3$$ , $$4a = 2$$

$$a = \frac{1}{2}$$     A1     N2

[4 marks]

c.

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

correct derivative of $$g$$     A1A1

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

evidence of equating both derivatives     (M1)

eg   $$f’ = g’$$

correct equation     (A1)

eg   $$\cos x + x – 2 = x – 2$$

working towards a solution     (A1)

eg   $$\cos x = 0$$ , combining like terms

$$x = \frac{\pi }{2}$$    A1     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 passes through $$(1, 7)$$.

Write down the value of $$h$$ and of $$k$$.

[2]
a.

Find the value of $$a$$.

[3]
b.

## Markscheme

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

[2 marks]

a.

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

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

correct equation     (A1)

eg     $$7 = a + 3$$

a = 4     A1     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 $$f$$ 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 $$36 – 12p$$     A2     N2

[2 marks]

a(i).

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

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

$$p = 3$$     AG     N0

[1 mark]

a(ii).

METHOD 1

valid approach     (M1)

eg     $$x = – \frac{b}{{2a}}$$

correct working     A1

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

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

METHOD 2

valid approach     (M1)

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

correct working     A1

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

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

METHOD 3

valid approach using derivative     (M1)

eg     $$f'(x) = 0,{\text{ }}6x – 6$$

correct equation     A1

eg     $$6x – 6 = 0$$

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

[4 marks]

b.

$$x = 1$$     A1     N1

[1 mark]

c.

$$a = 3$$     A1     N1

[1 mark]

d(i).

$$h = 1$$     A1     N1

[1 mark]

d(ii).

$$k = 0$$     A1     N1

[1 mark]

d(iii).

attempt to apply vertical reflection     (M1)

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

attempt to apply vertical shift 6 units up     (M1)

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

transformations performed correctly (in correct order)     (A1)

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

$$g(x) = – 3{x^2} + 6x + 3$$     A1     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 $$– 6,{\text{ }}(0,{\text{ }} – 6),{\text{ }}y = – 6$$     A1

[1 mark]

a.

valid attempt to solve     (M1)

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

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

[3 marks]

b.

A1A1A1

Note:     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 $$x = – \frac{1}{2}$$, below $$y = – 6$$,

A1 for both the $$x$$-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 $$x = – 1.5$$.

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 $$– 1.5$$ (seen anywhere)     (M1)

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

correct working to find the zeroes     A1

eg$$\,\,\,\,\,$$$$– 1.5 \pm 4.5$$

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

[2 marks]

a.

METHOD 1 (using factors)

attempt to write factors     (M1)

eg$$\,\,\,\,\,$$$$(x – 6)(x + 3)$$

correct factors     A1

eg$$\,\,\,\,\,$$$$(x – 3)(x + 6)$$

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

METHOD 2 (using derivative or vertex)

valid approach to find $$q$$     (M1)

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

$$q = 3$$    A1

correct substitution     A1

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

$$r = – 18$$    A1

$$q = 3,{\text{ }}r = – 18$$    N3

METHOD 3 (solving simultaneously)

valid approach setting up system of two equations     (M1)

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

one correct value

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

correct substitution     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     A1

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

$$q = 3,{\text{ }}r = – 18$$    N3

[4 marks]

b.

## Examiners report

As a ‘show that’ 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 − 4xc. A horizontal line, L , intersects the graph of f at x = −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 find p)

valid approach      (M1)

eg  $$\frac{{ – 1 + 3}}{2}$$,

p = 1     A1 N2

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

Do not accept $$p = \frac{2}{a}$$.

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

eg   a + 4 − c = a(32) − 4(3) − c,  f(−1) = f(3)

correct working      A1

eg   8a = 16

a = 2      AG N0

valid approach to find p      (M1)

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

p = 1      A1 N2

[2 marks]

a.i.

METHOD 1

valid approach       M1

eg  $$– \frac{b}{{2a}},\,\,\,\frac{4}{{2a}}$$ (might be seen in (i)), f’ (1) = 0

correct equation     A1

eg  $$\frac{4}{{2a}}$$ = 1, 2a(1) − 4 = 0

a = 2      AG N0

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

eg   a + 4 − c = a(32) − 4(3) − c,  f(−1) = f(3)

correct working      A1

eg   8a = 16

a = 2      AG N0

[2 marks]

a.ii.

valid approach      (M1)
eg   f(−1) = 5, f(3) =5

correct working       (A1)
eg   2 + 4 − c = 5, 18 − 12 − c = 5

c = 1     A1 N2

[3 marks]

b.

## Question

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

## Markscheme

METHOD 1

evidence of discriminant      (M1)
eg  $${b^2} – 4ac,\,\,\Delta$$

correct substitution into discriminant      (A1)
eg  $${q^2} – 4p\left( { – 4p} \right)$$

correct discriminant       A1
eg  $${q^2} + 16{p^2}$$

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

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

$${q^2} + 16{p^2} > 0$$      A1

$$f$$ has 2 roots     A1 N0

METHOD 2

y-intercept = −4p (seen anywhere)      A1

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

an upward-opening parabola with a negative y-intercept      R1
eg  sketch that must indicate p > 0.

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

a downward-opening parabola with a positive y-intercept      R1
eg  sketch that must indicate p > 0.

$$f$$ has 2 roots     A2 N0

[7 marks]

## 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)     write down the coordinates of the vertex;

(ii)    write down the y-intercept;

(iii)   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

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

$$= 3{x^2} + 6x + 3 – 12$$     A1

$$= 3{x^2} + 6x – 9$$     AG     N0

[2 marks]

a.

(i) vertex is $$( – 1, – 12)$$     A1A1     N2

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

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

e.g. factorizing, formula

correct working     A1

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

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

[7 marks]

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

A1A1A1     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.

$$\left( \begin{array}{l} p\\ q \end{array} \right) = \left( \begin{array}{l} – 1\\ – 12 \end{array} \right)$$ , $$t = 3$$     A1A1A1     N3

[3 marks]

d.