IB DP Maths Topic 2.3 Transformations of graphs SL Paper 2

 

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Question

Consider the graph of \(f\) shown below.


The following four diagrams show images of f under different transformations.


On the same grid sketch the graph of \(y = f( – x)\) .

[2]
a.

Complete the following table.


[2]
b.

Give a full geometric description of the transformation that gives the image in Diagram A.

[2]
c.
Answer/Explanation

Markscheme


     A2     N2

[2 marks]

a.

     A1A1     N2

[2 marks]

b.

translation (accept move/shift/slide etc.) with vector \(\left( {\begin{array}{*{20}{c}}
{ – 6}\\
{ – 2}
\end{array}} \right)\)     A1A1     N2

[2 marks]

c.

Question

Let \(f(x) = 3{x^2}\) . The graph of f is translated 1 unit to the right and 2 units down. The graph of g is the image of the graph of f after this translation.

Write down the coordinates of the vertex of the graph of g .

[2]
a.

Express g in the form \(g(x) = 3{(x – p)^2} + q\) .

[2]
b.

The graph of h is the reflection of the graph of g in the x-axis.

Write down the coordinates of the vertex of the graph of h .

[2]
c.
Answer/Explanation

Markscheme

\((1{\text{, }} – 2)\)     A1A1     N2

[2 marks]

a.

\(g(x) = 3{(x – 1)^2} – 2\) (accept \(p = 1\) , \(q = – 2\) )     A1A1     N2

[2 marks]

b.

\((1{\text{, }}2)\)     A1A1     N2

[2 marks]

c.

Question

Let \(f(t) = 2{t^2} + 7\) , where \(t > 0\) . The function v is obtained when the graph of f is transformed by

a stretch by a scale factor of \(\frac{1}{3}\) parallel to the y-axis,

followed by a translation by the vector \(\left( {\begin{array}{*{20}{c}}
2\\
{ – 4}
\end{array}} \right)\) .

Find \(v(t)\) , giving your answer in the form \(a{(t – b)^2} + c\) .

[4]
a.

A particle moves along a straight line so that its velocity in ms−1 , at time t seconds, is given by v . Find the distance the particle travels between \(t = 5.0\) and \(t = 6.8\) .

[3]
b.
Answer/Explanation

Markscheme

applies vertical stretch parallel to the y-axis factor of \(\frac{1}{3}\)     (M1)

e.g. multiply by \(\frac{1}{3}\) , \(\frac{1}{3}f(t)\) , \(\frac{1}{3} \times 2\)

applies horizontal shift 2 units to the right     (M1)

e.g. \(f(t – 2)\) , \(t – 2\)

applies a vertical shift 4 units down     (M1)

e.g. subtracting 4, \(f(t) – 4\) , \(\frac{7}{3} – 4\)

\(v(t) = \frac{2}{3}{(t – 2)^2} – \frac{5}{3}\)     A1     N4

[4 marks]

a.

recognizing that distance travelled is area under the curve     M1

e.g. \(\int {v,\frac{2}{9}} {(t – 2)^3} – \frac{5}{3}t\) , sketch


distance = 15.576 (accept 15.6)     A2     N2

[3 marks]

b.

Question

Consider the function \(f(x) = {x^2} – 4x + 1\) .

Sketch the graph of f , for \( – 1 \le x \le 5\) .

[4]
a.

This function can also be written as \(f(x) = {(x – p)^2} – 3\) .

Write down the value of p .

[1]
b.

The graph of g is obtained by reflecting the graph of f in the x-axis, followed by a translation of \(\left( {\begin{array}{*{20}{c}}
0\\
6
\end{array}} \right)\) .

Show that \(g(x) = – {x^2} + 4x + 5\) .  

[4]
c.

The graph of g is obtained by reflecting the graph of f in the x-axis, followed by a translation of \(\left( {\begin{array}{*{20}{c}}
0\\
6
\end{array}} \right)\) .

The graphs of f and g intersect at two points.

Write down the x-coordinates of these two points.

[3]
d.

The graph of \(g\) is obtained by reflecting the graph of \(f\) in the x-axis, followed by a translation of \(\left( {\begin{array}{*{20}{c}}
  0 \\
  6
\end{array}} \right)\) .

Let R be the region enclosed by the graphs of f and g .

Find the area of R .

[3]
e.
Answer/Explanation

Markscheme


     A1A1A1A1     N4

Note: The shape must be an approximately correct upwards parabola.

Only if the shape is approximately correct, award the following:

A1 for vertex \(x \approx 2\) , A1 for x-intercepts between 0 and 1, and 3 and 4, A1 for correct y-intercept \((0{\text{, }}1)\), A1 for correct domain \([ – 1{\text{, }}5]\).

Scale not required on the axes, but approximate positions need to be clear.

[4 marks]

a.

\(p = 2\)     A1     N1 

[1 mark]

b.

correct vertical reflection, correct vertical translation     (A1)(A1)

e.g. \( – f(x)\) , \( – ({(x – 2)^2} – 3)\) , \( – y\) , \( – f(x) + 6\) , \(y + 6\) 

transformations in correct order     (A1)

e.g. \( – ({x^2} – 4x + 1) + 6\) , \( – ({(x – 2)^2} – 3) + 6\)

simplification which clearly leads to given answer     A1

e.g. \( – {x^2} + 4x – 1 + 6\) , \( – ({x^2} – 4x + 4 – 3) + 6\)

\(g(x) = – {x^2} + 4x + 5\)     AG     N0

Note: If working shown, award A1A1A0A0 if transformations correct, but done in reverse order, e.g. \( – ({x^2} – 4x + 1 + 6)\).

[4 marks]

c.

valid approach     (M1)

e.g. sketch, \(f = g\)

\( – 0.449489 \ldots \) , \(4.449489 \ldots \)

\((2 \pm \sqrt 6 )\) (exact), \( – 0.449{\text{ }}[ – 0.450{\text{, }} – 0.449]\) ; \(4.45{\text{ }}[4.44{\text{, }}4.45]\)     A1A1     N3

[3 marks]

d.

attempt to substitute limits or functions into area formula (accept absence of \({\rm{d}}x\) )     (M1)

e.g. \(\int_a^b {(( – {x^2}}  + 4x + 5) – ({x^2} – 4x + 1)){\rm{d}}x\) , \(\int_{4.45}^{ – 0.449} {(f – g)} \) , \(\int {( – 2{x^2}}  + 8x + 4){\rm{d}}x\)

approach involving subtraction of integrals/areas (accept absence of \({\rm{d}}x\) )     (M1)

e.g. \(\int_a^b {( – {x^2}}  + 4x + 5) – \int_a^b {({x^2}}  – 4x + 1)\) , \(\int {(f – g){\rm{d}}x} \)

\({\rm{area}} = 39.19183 \ldots \)

\({\rm{area}} = 39.2\) \([39.1{\text{, }}39.2]\)     A1     N3

[3 marks]

e.

Question

Let \(f\) and \(g\) be functions such that \(g(x) = 2f(x + 1) + 5\) .

(a)     The graph of \(f\) is mapped to the graph of \(g\) under the following transformations:

vertical stretch by a factor of \(k\) , followed by a translation \(\left( \begin{array}{l}
p\\
q
\end{array} \right)\)
.

Write down the value of

  (i)     \(k\) ;

  (ii)     \(p\) ;

  (iii)     \(q\) .

(b)     Let \(h(x) = – g(3x)\) . The point A(\(6\), \(5\)) on the graph of \(g\) is mapped to the point \({\rm{A}}’\) on the graph of \(h\) . Find \({\rm{A}}’\) .

[6]
.

The graph of \(f\) is mapped to the graph of \(g\) under the following transformations:

vertical stretch by a factor of \(k\) , followed by a translation \(\left( \begin{array}{l}
p\\
q
\end{array} \right)\)
.

Write down the value of

  (i)     \(k\) ;

  (ii)     \(p\) ;

  (iii)     \(q\) .

[3]
a.

Let \(h(x) = – g(3x)\) . The point A(\(6\), \(5\)) on the graph of \(g\) is mapped to the point \({\rm{A}}’\) on the graph of \(h\) . Find \({\rm{A}}’\) .

[3]
b.
Answer/Explanation

Markscheme

(a)     (i)     \(k = 2\)     A1     N1

(ii)     \(p = – 1\)     A1     N1

(iii)     \(q = 5\)     A1     N1

[3 marks]


(b)     recognizing one transformation      (M1)

eg   horizontal stretch by \(\frac{1}{3}\) , reflection in \(x\)-axis

\({\rm{A’}}\) is (\(2\), \( – 5\))     A1A1     N3

[3 marks]

 

Total [6 marks]

.

(i)     \(k = 2\)     A1     N1

(ii)     \(p = – 1\)     A1     N1

(iii)     \(q = 5\)     A1     N1

[3 marks]

a.

recognizing one transformation      (M1)

eg   horizontal stretch by \(\frac{1}{3}\) , reflection in \(x\)-axis

\({\rm{A’}}\) is (\(2\), \( – 5\))     A1A1     N3

[3 marks]

 

Total [6 marks]

b.

Question

Let \(f(x) = \ln x\) and \(g(x) = 3 + \ln \left( {\frac{x}{2}} \right)\), for \(x > 0\).

The graph of \(g\) can be obtained from the graph of \(f\) by two transformations:

\[\begin{array}{*{20}{l}} {{\text{a horizontal stretch of scale factor }}q{\text{ followed by}}} \\ {{\text{a translation of }}\left( {\begin{array}{*{20}{c}} h \\ k \end{array}} \right).} \end{array}\]

Let \(h(x) = g(x) \times \cos (0.1x)\), for \(0 < x < 4\). The following diagram shows the graph of \(h\) and the line \(y = x\).

M17/5/MATME/SP2/ENG/TZ1/10.b.c

The graph of \(h\) intersects the graph of \({h^{ – 1}}\) at two points. These points have \(x\) coordinates 0.111 and 3.31 correct to three significant figures.

Write down the value of \(q\);

[1]
a.i.

Write down the value of \(h\);

[1]
a.ii.

Write down the value of \(k\).

[1]
a.iii.

Find \(\int_{0.111}^{3.31} {\left( {h(x) – x} \right){\text{d}}x} \).

[2]
b.i.

Hence, find the area of the region enclosed by the graphs of \(h\) and \({h^{ – 1}}\).

[3]
b.ii.

Let \(d\) be the vertical distance from a point on the graph of \(h\) to the line \(y = x\). There is a point \({\text{P}}(a,{\text{ }}b)\) on the graph of \(h\) where \(d\) is a maximum.

Find the coordinates of P, where \(0.111 < a < 3.31\).

[7]
c.
Answer/Explanation

Markscheme

\(q = 2\)     A1     N1

Note:     Accept \(q = 1\), \(h = 0\), and \(k = 3 – \ln (2)\), 2.31 as candidate may have rewritten \(g(x)\) as equal to \(3 + \ln (x) – \ln (2)\).

[1 mark]

a.i.

\(h = 0\)     A1     N1

Note:     Accept \(q = 1\), \(h = 0\), and \(k = 3 – \ln (2)\), 2.31 as candidate may have rewritten \(g(x)\) as equal to \(3 + \ln (x) – \ln (2)\).

[1 mark]

a.ii.

\(k = 3\)     A1     N1

Note:     Accept \(q = 1\), \(h = 0\), and \(k = 3 – \ln (2)\), 2.31 as candidate may have rewritten \(g(x)\) as equal to \(3 + \ln (x) – \ln (2)\).

[1 mark]

a.iii.

2.72409

2.72     A2     N2

[2 marks]

b.i.

recognizing area between \(y = x\) and \(h\) equals 2.72     (M1)

eg\(\,\,\,\,\,\)M17/5/MATME/SP2/ENG/TZ1/10.b.ii/M

recognizing graphs of \(h\) and \({h^{ – 1}}\) are reflections of each other in \(y = x\)     (M1)

eg\(\,\,\,\,\,\)area between \(y = x\) and \(h\) equals between \(y = x\) and \({h^{ – 1}}\)

\(2 \times 2.72\int_{0.111}^{3.31} {\left( {x – {h^{ – 1}}(x)} \right){\text{d}}x = 2.72} \)

5.44819

5.45     A1     N3

[??? marks]

b.ii.

valid attempt to find \(d\)     (M1)

eg\(\,\,\,\,\,\)difference in \(y\)-coordinates, \(d = h(x) – x\)

correct expression for \(d\)     (A1)

eg\(\,\,\,\,\,\)\(\left( {\ln \frac{1}{2}x + 3} \right)(\cos 0.1x) – x\)

valid approach to find when \(d\) is a maximum     (M1)

eg\(\,\,\,\,\,\)max on sketch of \(d\), attempt to solve \(d’ = 0\)

0.973679

\(x = 0.974\)     A2     N4 

substituting their \(x\) value into \(h(x)\)     (M1)

2.26938

\(y = 2.27\)     A1     N2

[7 marks]

c.
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