# IB Math Analysis & Approaches Questionbank-Topic: SL 2.11 Transformations of graphs SL Paper 1

## 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

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

The vertex is at $$(1,{\text{ }} – 9)$$, and the graph crosses the yaxis at the point $$(0,{\text{ }}c)$$.

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

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

[2]
a.

Find the value of $$c$$.

[2]
b.

Let $$g(x) = – {(x – 3)^2} + 1$$. The graph of $$g$$ is obtained by a reflection of the graph of $$f$$ in the $$x$$-axis, followed by a translation of $$\left( {\begin{array}{*{20}{c}} p \\ q \end{array}} \right)$$.

Find the value of $$p$$ and of $$q$$.

[5]
c.

Find the x-coordinates of the points of intersection of the graphs of $$f$$ and $$g$$.

[7]
d.

## Markscheme

$$h = 1,{\text{ }}k = – 9\;\;\;\left( {{\text{accept }}{{(x – 1)}^2} – 9} \right)$$     A1A1     N2

[2 marks]

a.

METHOD 1

attempt to substitute $$x = 0$$ into their quadratic function     (M1)

eg$$\;\;\;f(0),{\text{ }}{(0 – 1)^2} – 9$$

$$c = – 8$$     A1     N2

METHOD 2

attempt to expand their quadratic function     (M1)

eg$$\;\;\;{x^2} – 2x + 1 – 9,{\text{ }}{x^2} – 2x – 8$$

$$c = – 8$$     A1     N2

[2 marks]

b.

evidence of correct reflection     A1

eg$$\;\;\; – \left( {{{(x – 1)}^2} – 9} \right)$$, vertex at $$(1,{\text{ }}9)$$, y-intercept at $$(0,{\text{ }}8)$$

valid attempt to find horizontal shift     (M1)

eg$$\;\;\;1 + p = 3,{\text{ }}1 \to 3$$

$$p = 2$$     A1     N2

valid attempt to find vertical shift     (M1)

eg$$\;\;\;9 + q = 1,{\text{ }}9 \to 1,{\text{ }} – 9 + q = 1$$

$$q = – 8$$     A1     N2

Notes:     An error in finding the reflection may still allow the correct values of $$p$$ and $$q$$ to be found, as the error may not affect subsequent working. In this case, award A0 for the reflection, M1A1 for $$p = 2$$, and M1A1 for $$q = – 8$$.

If no working shown, award N0 for $$q = 10$$.

[5 marks]

c.

valid approach (check FT from (a))     M1

eg$$\;\;\;f(x) = g(x),{\text{ }}{(x – 1)^2} – 9 = – {(x – 3)^2} + 1$$

correct expansion of both binomials     (A1)

eg$$\;\;\;{x^2} – 2x + 1,{\text{ }}{x^2} – 6x + 9$$

correct working     (A1)

eg$$\;\;\;{x^2} – 2x – 8 = – {x^2} + 6x – 8$$

correct equation     (A1)

eg$$\;\;\;2{x^2} – 8x = 0,{\text{ }}2{x^2} = 8x$$

correct working     (A1)

eg$$\;\;\;2x(x – 4) = 0$$

$$x = 0,{\text{ }}x = 4$$     A1A1     N3

[7 marks]

Total [16 marks]

d.

## Question

The following diagram shows the graph of a function $$f$$, for −4 ≤ x ≤ 2.

On the same axes, sketch the graph of $$f\left( { – x} \right)$$.

[2]
a.

Another function, $$g$$, can be written in the form $$g\left( x \right) = a \times f\left( {x + b} \right)$$. The following diagram shows the graph of $$g$$.

Write down the value of a and of b.

[4]
b.

## Markscheme

A2 N2
[2 marks]

a.

recognizing horizontal shift/translation of 1 unit      (M1)

eg  = 1, moved 1 right

recognizing vertical stretch/dilation with scale factor 2      (M1)

eg   a = 2,  ×(−2)

a = −2,  b = −1     A1A1 N2N2

[4 marks]

b.

## Question

The following diagram shows the graph of a function $$f$$.

Find $${f^{ – 1}}( – 1)$$.

[2]
a.

Find $$(f \circ f)( – 1)$$.

[3]
b.

On the same diagram, sketch the graph of $$y = f( – x)$$.

[2]
c.

## Markscheme

valid approach     (M1)

eg$$\;\;\;$$horizontal line on graph at $$– 1,{\text{ }}f(a) = – 1,{\text{ }}( – 1,5)$$

$${f^{ – 1}}( – 1) = 5$$     A1     N2

[2 marks]

a.

attempt to find $$f( – 1)$$     (M1)

eg$$\;\;\;$$line on graph

$$f( – 1) = 2$$     (A1)

$$(f \circ f)( – 1) = 1$$     A1     N3

[3 marks]

b.

A1A1     N2

Note:     The shape must be an approximately correct shape (concave down and increasing). Only if the shape is approximately correct, award the following for points in circles:

A1 for the $$y$$-intercept,

A1 for any two of these points $$( – 5,{\text{ }} – 1),{\text{ }}( – 2,{\text{ }}1),{\text{ }}(1,{\text{ }}2)$$.

[2 marks]

Total [7 marks]

c.

## 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

Part of the graph of a function f is shown in the diagram below.

On the same diagram sketch the graph of $$y = – f(x)$$ .

[2]
a.

Let $$g(x) = f(x + 3)$$ .

(i)     Find $$g( – 3)$$ .

(ii)    Describe fully the transformation that maps the graph of f to the graph of g.

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

## Markscheme

M1A1     N2

Note: Award M1 for evidence of reflection in x-axis, A1 for correct vertex and all intercepts approximately correct.

a.

(i) $$g( – 3) = f(0)$$     (A1)

$$f(0) = – 1.5$$     A1     N2

(ii) translation (accept shift, slide, etc.) of $$\left( {\begin{array}{*{20}{c}} { – 3}\\ 0 \end{array}} \right)$$
A1A1     N2

[4 marks]

b(i) and (ii).

## Question

Let $$f(t) = a\cos b(t – c) + d$$ , $$t \ge 0$$ . Part of the graph of $$y = f(t)$$ is given below.

When $$t = 3$$ , there is a maximum value of 29, at M.

When $$t = 9$$ , there is a minimum value of 15.

(i)     Find the value of a.

(ii)    Show that $$b = \frac{\pi }{6}$$ .

(iii)   Find the value of d.

(iv)   Write down a value for c.

[7]
a(i), (ii), (iii) and (iv).

The transformation P is given by a horizontal stretch of a scale factor of $$\frac{1}{2}$$ , followed by a translation of $$\left( {\begin{array}{*{20}{c}} 3\\ { – 10} \end{array}} \right)$$ .

Let $${M’}$$ be the image of M under P. Find the coordinates of $${M’}$$ .

[2]
b.

The graph of g is the image of the graph of f under P.

Find $$g(t)$$ in the form $$g(t) = 7\cos B(t – c) + D$$ .

[4]
c.

The graph of g is the image of the graph of f under P.

Give a full geometric description of the transformation that maps the graph of g to the graph of f .

[3]
d.

## Markscheme

(i) attempt to substitute     (M1)

e.g. $$a = \frac{{29 – 15}}{2}$$

$$a = 7$$ (accept $$a = – 7$$ )     A1     N2

(ii) $${\text{period}} = 12$$     (A1)

$$b = \frac{{2\pi }}{{12}}$$    A1

$$b = \frac{\pi }{6}$$    AG     N0

(iii) attempt to substitute     (M1)

e.g. $$d = \frac{{29 + 15}}{2}$$

$$d = 22$$     A1     N2

(iv) $$c = 3$$ (accept $$c = 9$$ from $$a = – 7$$ )     A1     N1

Note: Other correct values for c can be found, $$c = 3 \pm 12k$$ , $$k \in \mathbb{Z}$$ .

[7 marks]

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

stretch takes 3 to 1.5     (A1)

translation maps $$(1.5{\text{, }}29)$$ to $$(4.5{\text{, }}19)$$ (so $${M’}$$ is $$(4.5{\text{, }}19)$$)     A1     N2

[2 marks]

b.

$$g(t) = 7\cos \frac{\pi }{3}\left( {t – 4.5} \right) + 12$$    A1A2A1    N4

Note: Award A1 for $$\frac{\pi }{3}$$ , A2 for 4.5, A1 for 12.

Other correct values for c can be found, $$c = 4.5 \pm 6k$$ , $$k \in \mathbb{Z}$$ .

[4 marks]

c.

translation $$\left( {\begin{array}{*{20}{c}} { – 3}\\ {10} \end{array}} \right)$$     (A1)

horizontal stretch of a scale factor of 2     (A1)

completely correct description, in correct order     A1     N3

e.g. translation $$\left( {\begin{array}{*{20}{c}} { – 3}\\ {10} \end{array}} \right)$$ then horizontal stretch of a scale factor of 2

[3 marks]

d.

## Question

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

The graph of g can be obtained from the graph of f using two transformations.

Give a full geometric description of each of the two transformations.

[2]
a.

The graph of g is translated by the vector $$\left( {\begin{array}{*{20}{c}} 3\\ { – 2} \end{array}} \right)$$ to give the graph of h.

The point $$( – 1{\text{, }}1)$$ on the graph of f is translated to the point P on the graph of h.

Find the coordinates of P.

[4]
b.

## Markscheme

in any order

translated 1 unit to the right     A1     N1

stretched vertically by factor 2     A1     N1

[2 marks]

a.

METHOD 1

finding coordinates of image on g     (A1)(A1)

e.g.  $$– 1 + 1 = 0$$ , $$1 \times 2 = 2$$ , $$( – 1{\text{, }}1) \to ( – 1 + 1{\text{, }}2 \times 1)$$ , $$(0{\text{, }}2)$$

P is (3, 0)     A1A1     N4

METHOD 2

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

P is $$(3{\text{, }}0)$$     A1A1     N4

b.

## Question

Let $$f(x) = \frac{{ax}}{{{x^2} + 1}}$$ , $$– 8 \le x \le 8$$ , $$a \in \mathbb{R}$$ .The graph of f is shown below.

The region between $$x = 3$$ and $$x = 7$$ is shaded.

Show that $$f( – x) = – f(x)$$ .

[2]
a.

Given that $$f”(x) = \frac{{2ax({x^2} – 3)}}{{{{({x^2} + 1)}^3}}}$$ , find the coordinates of all points of inflexion.

[7]
b.

It is given that $$\int {f(x){\rm{d}}x = \frac{a}{2}} \ln ({x^2} + 1) + C$$ .

(i)     Find the area of the shaded region, giving your answer in the form $$p\ln q$$ .

(ii)    Find the value of $$\int_4^8 {2f(x – 1){\rm{d}}x}$$ .

[7]
c.

## Markscheme

METHOD 1

evidence of substituting $$– x$$ for $$x$$     (M1)

$$f( – x) = \frac{{a( – x)}}{{{{( – x)}^2} + 1}}$$     A1

$$f( – x) = \frac{{ – ax}}{{{x^2} + 1}}$$ $$( = – f(x))$$     AG     N0

METHOD 2

$$y = – f(x)$$ is reflection of $$y = f(x)$$ in x axis

and $$y = f( – x)$$ is reflection of $$y = f(x)$$ in y axis     (M1)

sketch showing these are the same     A1

$$f( – x) = \frac{{ – ax}}{{{x^2} + 1}}$$ $$( = – f(x))$$     AG     N0

[2 marks]

a.

evidence of appropriate approach     (M1)

e.g. $$f”(x) = 0$$

to set the numerator equal to 0     (A1)

e.g. $$2ax({x^2} – 3) = 0$$ ; $$({x^2} – 3) = 0$$

(0, 0) , $$\left( {\sqrt 3 ,\frac{{a\sqrt 3 }}{4}} \right)$$ , $$\left( { – \sqrt 3 , – \frac{{a\sqrt 3 }}{4}} \right)$$ (accept $$x = 0$$ , $$y = 0$$ etc)      A1A1A1A1A1     N5

[7 marks]

b.

(i) correct expression     A2

e.g. $$\left[ {\frac{a}{2}\ln ({x^2} + 1)} \right]_3^7$$ , $$\frac{a}{2}\ln 50 – \frac{a}{2}\ln 10$$ , $$\frac{a}{2}(\ln 50 – \ln 10)$$

area = $$\frac{a}{2}\ln 5$$     A1A1     N2

(ii) METHOD 1

recognizing the shift that does not change the area     (M1)

e.g. $$\int_4^8 {f(x – 1){\rm{d}}x} = \int_3^7 {f(x){\rm{d}}x}$$ , $$\frac{a}{2}\ln 5$$

recognizing that the factor of 2 doubles the area     (M1)

e.g. $$\int_4^8 {2f(x – 1){\rm{d}}x = } 2\int_4^8 {f(x – 1){\rm{d}}x}$$ $$\left( { = 2\int_3^7 {f(x){\rm{d}}x} } \right)$$

$$\int_4^8 {2f(x – 1){\rm{d}}x = a\ln 5}$$ (i.e. $$2 \times$$ their answer to (c)(i))     A1     N3

METHOD 2

changing variable

let $$w = x – 1$$ , so $$\frac{{{\rm{d}}w}}{{{\rm{d}}x}} = 1$$

$$2\int {f(w){\rm{d}}w = } \frac{{2a}}{2}\ln ({w^2} + 1) + c$$     (M1)

substituting correct limits

e.g. $$\left[ {a\ln \left[ {{{(x – 1)}^2} + 1} \right]} \right]_4^8$$ , $$\left[ {a\ln ({w^2} + 1)} \right]_3^7$$ , $$a\ln 50 – a\ln 10$$     (M1)

$$\int_4^8 {2f(x – 1){\rm{d}}x = a\ln 5}$$     A1     N3

[7 marks]

c.

## Question

The diagram below shows the graph of a function $$f(x)$$ , for $$– 2 \le x \le 4$$ .

Let $$h(x) = f( – x)$$ . Sketch the graph of $$h$$ on the grid below.

[3]
a.

Let $$g(x) = \frac{1}{2}f(x – 1)$$ . The point $${\text{A}}(3{\text{, }}2)$$ on the graph of $$f$$ is transformed to the point P on the graph of $$g$$ . Find the coordinates of P.

[3]
b.

## Markscheme

A2     N2

[2 marks]

a.

evidence of appropriate approach     (M1)

e.g. reference to any horizontal shift and/or stretch factor, $$x = 3 + 1$$ , $$y = \frac{1}{2} \times 2$$

P is $$(4{\text{, }}1)$$ (accept $$x = 4$$ , $$y = 1$$)     A1A1     N3

[3 marks]

b.

## Question

Let $$f(x) = \frac{1}{2}{x^3} – {x^2} – 3x$$ . Part of the graph of f is shown below.

There is a maximum point at A and a minimum point at B(3, − 9) .

Find the coordinates of A.

[8]
a.

Write down the coordinates of

(i)     the image of B after reflection in the y-axis;

(ii)    the image of B after translation by the vector $$\left( {\begin{array}{*{20}{c}} { – 2}\\ 5 \end{array}} \right)$$ ;

(iii)   the image of B after reflection in the x-axis followed by a horizontal stretch with scale factor $$\frac{1}{2}$$ .

[6]
b(i), (ii) and (iii).

## Markscheme

$$f(x) = {x^2} – 2x – 3$$     A1A1A1

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

e.g. $${x^2} – 2x – 3 = 0$$

evidence of correct working     A1

e.g. $$(x + 1)(x – 3)$$ ,  $$\frac{{2 \pm \sqrt {16} }}{2}$$

$$x = – 1$$ (ignore $$x = 3$$ )     (A1)

evidence of substituting their negative x-value into $$f(x)$$     (M1)

e.g. $$\frac{1}{3}{( – 1)^3} – {( – 1)^2} – 3( – 1)$$ , $$– \frac{1}{3} – 1 + 3$$

$$y = \frac{5}{3}$$     A1

coordinates are $$\left( { – 1,\frac{5}{3}} \right)$$     N3

[8 marks]

a.

(i) $$( – 3{\text{, }} – 9)$$     A1     N1

(ii) $$(1{\text{, }} – 4)$$     A1A1    N2

(iii) reflection gives $$(3{\text{, }}9)$$     (A1)

stretch gives $$\left( {\frac{3}{2}{\text{, }}9} \right)$$     A1A1     N3

[6 marks]

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

## 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.

## Question

Let $$f(x) = 3\ln x$$ and $$g(x) = \ln 5{x^3}$$ .

Express $$g(x)$$ in the form $$f(x) + \ln a$$ , where $$a \in {{\mathbb{Z}}^ + }$$ .

[4]
a.

The graph of g is a transformation of the graph of f . Give a full geometric description of this transformation.

[3]
b.

## Markscheme

attempt to apply rules of logarithms     (M1)

e.g. $$\ln {a^b} = b\ln a$$ , $$\ln ab = \ln a + \ln b$$

correct application of $$\ln {a^b} = b\ln a$$ (seen anywhere)     A1

e.g. $$3\ln x = \ln {x^3}$$

correct application of $$\ln ab = \ln a + \ln b$$ (seen anywhere)     A1

e.g. $$\ln 5{x^3} = \ln 5 + \ln {x^3}$$

so $$\ln 5{x^3} = \ln 5 + 3\ln x$$

$$g(x) = f(x) + \ln 5$$ (accept $$g(x) = 3\ln x + \ln 5$$ )     A1     N1

[4 marks]

a.

transformation with correct name, direction, and value     A3

e.g. translation by $$\left( {\begin{array}{*{20}{c}} 0\\ {\ln 5} \end{array}} \right)$$ , shift up by $$\ln 5$$ , vertical translation of $$\ln 5$$

[3 marks]

b.

## Examiners report

This question was very poorly done by the majority of candidates. While candidates seemed to have a vague idea of how to apply the rules of logarithms in part (a), very few did so successfully. The most common error in part (a) was to begin incorrectly with $$\ln 5{x^3} = 3\ln 5x$$ . This error was often followed by other errors.

a.

In part (b), very few candidates were able to describe the transformation as a vertical translation (or shift). Many candidates attempted to describe numerous incorrect transformations, and some left part (b) entirely blank.

b.

## Question

The diagram below shows the graph of a function $$f(x)$$ , for $$– 2 \le x \le 3$$ .

Sketch the graph of $$f( – x)$$ on the grid below.

[2]
a.

The graph of f is transformed to obtain the graph of g . The graph of g is shown below.

The function g can be written in the form $$g(x) = af(x + b)$$ . Write down the value of a and of b .

[4]
b.

## Markscheme

A2     N2

[2 marks]

a.

$$a = – 2,b = – 1$$     A2A2     N4

Note: Award A1 for $$a = 2$$ , A1 for $$b = 1$$ .

[4 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

The following diagram shows the graph of a function $$f$$.

Find $${f^{ – 1}}( – 1)$$.

[2]
a.

Find $$(f \circ f)( – 1)$$.

[3]
b.

On the same diagram, sketch the graph of $$y = f( – x)$$.

[2]
c.

## Markscheme

valid approach     (M1)

eg$$\;\;\;$$horizontal line on graph at $$– 1,{\text{ }}f(a) = – 1,{\text{ }}( – 1,5)$$

$${f^{ – 1}}( – 1) = 5$$     A1     N2

[2 marks]

a.

attempt to find $$f( – 1)$$     (M1)

eg$$\;\;\;$$line on graph

$$f( – 1) = 2$$     (A1)

$$(f \circ f)( – 1) = 1$$     A1     N3

[3 marks]

b.

A1A1     N2

Note:     The shape must be an approximately correct shape (concave down and increasing). Only if the shape is approximately correct, award the following for points in circles:

A1 for the $$y$$-intercept,

A1 for any two of these points $$( – 5,{\text{ }} – 1),{\text{ }}( – 2,{\text{ }}1),{\text{ }}(1,{\text{ }}2)$$.

[2 marks]

Total [7 marks]

c.

## Question

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

The vertex is at $$(1,{\text{ }} – 9)$$, and the graph crosses the yaxis at the point $$(0,{\text{ }}c)$$.

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

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

[2]
a.

Find the value of $$c$$.

[2]
b.

Let $$g(x) = – {(x – 3)^2} + 1$$. The graph of $$g$$ is obtained by a reflection of the graph of $$f$$ in the $$x$$-axis, followed by a translation of $$\left( {\begin{array}{*{20}{c}} p \\ q \end{array}} \right)$$.

Find the value of $$p$$ and of $$q$$.

[5]
c.

Find the x-coordinates of the points of intersection of the graphs of $$f$$ and $$g$$.

[7]
d.

## Markscheme

$$h = 1,{\text{ }}k = – 9\;\;\;\left( {{\text{accept }}{{(x – 1)}^2} – 9} \right)$$     A1A1     N2

[2 marks]

a.

METHOD 1

attempt to substitute $$x = 0$$ into their quadratic function     (M1)

eg$$\;\;\;f(0),{\text{ }}{(0 – 1)^2} – 9$$

$$c = – 8$$     A1     N2

METHOD 2

attempt to expand their quadratic function     (M1)

eg$$\;\;\;{x^2} – 2x + 1 – 9,{\text{ }}{x^2} – 2x – 8$$

$$c = – 8$$     A1     N2

[2 marks]

b.

evidence of correct reflection     A1

eg$$\;\;\; – \left( {{{(x – 1)}^2} – 9} \right)$$, vertex at $$(1,{\text{ }}9)$$, y-intercept at $$(0,{\text{ }}8)$$

valid attempt to find horizontal shift     (M1)

eg$$\;\;\;1 + p = 3,{\text{ }}1 \to 3$$

$$p = 2$$     A1     N2

valid attempt to find vertical shift     (M1)

eg$$\;\;\;9 + q = 1,{\text{ }}9 \to 1,{\text{ }} – 9 + q = 1$$

$$q = – 8$$     A1     N2

Notes:     An error in finding the reflection may still allow the correct values of $$p$$ and $$q$$ to be found, as the error may not affect subsequent working. In this case, award A0 for the reflection, M1A1 for $$p = 2$$, and M1A1 for $$q = – 8$$.

If no working shown, award N0 for $$q = 10$$.

[5 marks]

c.

valid approach (check FT from (a))     M1

eg$$\;\;\;f(x) = g(x),{\text{ }}{(x – 1)^2} – 9 = – {(x – 3)^2} + 1$$

correct expansion of both binomials     (A1)

eg$$\;\;\;{x^2} – 2x + 1,{\text{ }}{x^2} – 6x + 9$$

correct working     (A1)

eg$$\;\;\;{x^2} – 2x – 8 = – {x^2} + 6x – 8$$

correct equation     (A1)

eg$$\;\;\;2{x^2} – 8x = 0,{\text{ }}2{x^2} = 8x$$

correct working     (A1)

eg$$\;\;\;2x(x – 4) = 0$$

$$x = 0,{\text{ }}x = 4$$     A1A1     N3

[7 marks]

Total [16 marks]

d.

## Question

Let $$f'(x) = \frac{{6 – 2x}}{{6x – {x^2}}}$$, for $$0 < x < 6$$.

The graph of $$f$$ has a maximum point at P.

The $$y$$-coordinate of P is $$\ln 27$$.

Find the $$x$$-coordinate of P.

[3]
a.

Find $$f(x)$$, expressing your answer as a single logarithm.

[8]
b.

The graph of $$f$$ is transformed by a vertical stretch with scale factor $$\frac{1}{{\ln 3}}$$. The image of P under this transformation has coordinates $$(a,{\text{ }}b)$$.

Find the value of $$a$$ and of $$b$$, where $$a,{\text{ }}b \in \mathbb{N}$$.

[[N/A]]
c.

## Markscheme

recognizing $$f'(x) = 0$$     (M1)

correct working     (A1)

eg$$\,\,\,\,\,$$$$6 – 2x = 0$$

$$x = 3$$    A1     N2

[3 marks]

a.

evidence of integration     (M1)

eg$$\,\,\,\,\,$$$$\int {f’,{\text{ }}\int {\frac{{6 – 2x}}{{6x – {x^2}}}{\text{d}}x} }$$

using substitution     (A1)

eg$$\,\,\,\,\,$$$$\int {\frac{1}{u}{\text{d}}u}$$ where $$u = 6x – {x^2}$$

correct integral     A1

eg$$\,\,\,\,\,$$$$\ln (u) + c,{\text{ }}\ln (6x – {x^2})$$

substituting $$(3,{\text{ }}\ln 27)$$ into their integrated expression (must have $$c$$)     (M1)

eg$$\,\,\,\,\,$$$$\ln (6 \times 3 – {3^2}) + c = \ln 27,{\text{ }}\ln (18 – 9) + \ln k = \ln 27$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$c = \ln 27 – \ln 9$$

EITHER

$$c = \ln 3$$    (A1)

attempt to substitute their value of $$c$$ into $$f(x)$$     (M1)

eg$$\,\,\,\,\,$$$$f(x) = \ln (6x – {x^2}) + \ln 3$$     A1     N4

OR

attempt to substitute their value of $$c$$ into $$f(x)$$     (M1)

eg$$\,\,\,\,\,$$$$f(x) = \ln (6x – {x^2}) + \ln 27 – \ln 9$$

correct use of a log law     (A1)

eg$$\,\,\,\,\,$$$$f(x) = \ln (6x – {x^2}) + \ln \left( {\frac{{27}}{9}} \right),{\text{ }}f(x) = \ln \left( {27(6x – {x^2})} \right) – \ln 9$$

$$f(x) = \ln \left( {3(6x – {x^2})} \right)$$    A1     N4

[8 marks]

b.

$$a = 3$$    A1     N1

correct working     A1

eg$$\,\,\,\,\,$$$$\frac{{\ln 27}}{{\ln 3}}$$

correct use of log law     (A1)

eg$$\,\,\,\,\,$$$$\frac{{3\ln 3}}{{\ln 3}},{\text{ }}{\log _3}27$$

$$b = 3$$    A1     N2

[4 marks]

c.

## Question

The following diagram shows the graph of a function $$f$$, for −4 ≤ x ≤ 2.

On the same axes, sketch the graph of $$f\left( { – x} \right)$$.

[2]
a.

Another function, $$g$$, can be written in the form $$g\left( x \right) = a \times f\left( {x + b} \right)$$. The following diagram shows the graph of $$g$$.

Write down the value of a and of b.

[4]
b.

## Markscheme

A2 N2
[2 marks]

a.

recognizing horizontal shift/translation of 1 unit      (M1)

eg  = 1, moved 1 right

recognizing vertical stretch/dilation with scale factor 2      (M1)

eg   a = 2,  ×(−2)

a = −2,  b = −1     A1A1 N2N2

[4 marks]

b.

## 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.