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.
Answer/Explanation

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

correct answers     A1A1     N2

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

correct answers     A1A1     N2

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     \(6x – 6 = 0\)

correct answers     A1A1     N2

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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).
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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).
Answer/Explanation

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.
Answer/Explanation

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)

e.g. factorizing, quadratic formula

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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

correct answers     A1A1     N2

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

correct answers     A1A1     N2

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     \(6x – 6 = 0\)

correct answers     A1A1     N2

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.

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