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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 \(y = f( – x)\) .
Complete the following table.
Give a full geometric description of the transformation that gives the image in Diagram A.
Answer/Explanation
Markscheme
A2 N2
[2 marks]
A1A1 N2
[2 marks]
translation (accept move/shift/slide etc.) with vector \(\left( {\begin{array}{*{20}{c}}
{ – 6}\\
{ – 2}
\end{array}} \right)\) A1A1 N2
[2 marks]
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 .
Express g in the form \(g(x) = 3{(x – p)^2} + q\) .
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 .
Answer/Explanation
Markscheme
\((1{\text{, }} – 2)\) A1A1 N2
[2 marks]
\(g(x) = 3{(x – 1)^2} – 2\) (accept \(p = 1\) , \(q = – 2\) ) A1A1 N2
[2 marks]
\((1{\text{, }}2)\) A1A1 N2
[2 marks]
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 \(a{(t – b)^2} + c\) .
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 \(t = 6.8\) .
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]
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]
Question
Consider the function \(f(x) = {x^2} – 4x + 1\) .
Sketch the graph of f , for \( – 1 \le x \le 5\) .
This function can also be written as \(f(x) = {(x – p)^2} – 3\) .
Write down the value of p .
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 \(g(x) = – {x^2} + 4x + 5\) .
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.
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 .
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]
\(p = 2\) A1 N1
[1 mark]
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]
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]
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]
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 \({\rm{A}}’\) .
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) \(q\) .
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 \({\rm{A}}’\) .
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]
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]
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 \(y = x\).
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 \(q\);
Write down the value of \(h\);
Write down the value of \(k\).
Find \(\int_{0.111}^{3.31} {\left( {h(x) – x} \right){\text{d}}x} \).
Hence, find the area of the region enclosed by the graphs of \(h\) and \({h^{ – 1}}\).
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 \(0.111 < a < 3.31\).
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]
\(h = 0\) 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]
\(k = 3\) 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]
2.72409
2.72 A2 N2
[2 marks]
recognizing area between \(y = x\) and \(h\) equals 2.72 (M1)
eg\(\,\,\,\,\,\)
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]
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]