# IB DP Math AA Topic SL 2.11 :Transformations of graphs: translations; stretches; HL Paper 2

## Question

The graph of $$y = \ln (x)$$ is transformed into the graph of $$y = \ln \left( {2x + 1} \right)$$ .

a.Describe two transformations that are required to do this.[2]

b.Solve $$\ln \left( {2x + 1} \right) > 3\cos (x)$$, $$x \in [0,10]$$.[4]

## Markscheme

EITHER

translation of $$– \frac{1}{2}$$ parallel to the $$x$$-axis

stretch of a scale factor of $$\frac{1}{2}$$ parallel to the $$x$$-axis     A1A1

OR

stretch of a scale factor of $$\frac{1}{2}$$ parallel to the $$x$$-axis

translation of $$– 1$$ parallel to the $$x$$-axis     A1A1

Note: Accept clear alternative terminologies for either transformation.

[2 marks]

a.

EITHER

$$1.16 < x < 5.71 \cup 6.75 < x \leqslant 10$$     A1A1A1A1

OR

]$$1.16$$, $$5.71$$[  $$\cup$$  ]$$6.75$$, $$10$$]     A1A1A1A1

Note: Award A1 for 1 intersection value, A1 for the other 2, A1A1 for the intervals.

[6 marks]

## Question

Let $$f(x) = \ln x$$ . The graph of f is transformed into the graph of the function by a translation of $$\left( {\begin{array}{*{20}{c}} 3 \\ { – 2} \end{array}} \right)$$, followed by a reflection in the x-axis. Find an expression for $$g(x)$$, giving your answer as a single logarithm.

## Markscheme

$$h(x) = f(x – 3) – 2 = \ln (x – 3) – 2$$     (M1)(A1)

$$g(x) = -h(x) = 2 – \ln (x – 3)$$     M1

Note: Award M1 only if it is clear the effect of the reflection in the x-axis:

the expression is correct OR

there is a change of signs of the previous expression OR

there’s a graph or an explanation making it explicit

$$= \ln {{\text{e}}^2} – \ln (x – 3)$$     M1

$$= \ln \left( {\frac{{{{\text{e}}^2}}}{{x – 3}}} \right)$$     A1

[5 marks]

## Examiners report

This question was well attempted but many candidates could have scored better had they written down all the steps to obtain the final expression. In some cases, as the final expression was incorrect and the middle steps were missing, candidates scored just 1 mark. That could be a consequence of a small mistake, but the lack of working prevented them from scoring at least all method marks. Some candidates performed the transformations well but were not able to use logarithms properties to transform the answer and give it as a single logarithm.

## Question

The graph of $$y = \ln (5x + 10)$$ is obtained from the graph of $$y = \ln x$$ by a translation of $$a$$ units in the direction of the $$x$$-axis followed by a translation of $$b$$ units in the direction of the $$y$$-axis.

a.Find the value of $$a$$ and the value of $$b$$.[4]

b.The region bounded by the graph of $$y = \ln (5x + 10)$$, the $$x$$-axis and the lines $$x = {\text{e}}$$ and $$x = 2{\text{e}}$$, is rotated through $$2\pi$$ radians about the $$x$$-axis. Find the volume generated.[2]

## Markscheme

EITHER

$$y = \ln (x – a) + b = \ln (5x + 10)$$     (M1)

$$y = \ln (x – a) + \ln c = \ln (5x + 10)$$

$$y = \ln \left( {c(x – a)} \right) = \ln (5x + 10)$$     (M1)

OR

$$y = \ln (5x + 10) = \ln \left( {5(x + 2)} \right)$$     (M1)

$$y = \ln (5) + \ln (x + 2)$$     (M1)

THEN

$$a = – 2,{\text{ }}b = \ln 5$$     A1A1

Note:     Accept graphical approaches.

Note:     Accept $$a = 2,{\text{ }}b = 1.61$$

[4 marks]

a.

$$V = \pi {\int_e^{2e} {\left[ {\ln (5x + 10)} \right]} ^2}{\text{d}}x$$     (M1)

$$= 99.2$$     A1

[2 marks]

Total [6 marks]

b.

## Examiners report

[N/A]

a.

[N/A]

b.

### Question

The graph of y = 2x2 + 4x +7 is translated using the vector $$\begin{pmatrix} 2\\-1\end{pmatrix}$$. Find the equation of the translated graph, giving your answer in the form y = ax2 + bx c.

$$y=2(x-2)^2+4(x-2)+7-1$$ $$=2(x^2-4x+4)+4x-8+6$$ $$=2x^2-8x+8+4x-2$$ $$\Rightarrow y = 2x^2-4x+6$$