EdExcel Core 1
Sketching curves
Chapter Assessment
Do not use a calculator for this test.
1.
2.
Sketch the following graphs on separate diagrams.
(i) y = x³
[1]
(ii) y = (x + 1)³
[2]
(iii) y = x³ – 2
[2]
(i) Sketch the graph of y = f(x), where f ( x)  ( x  1)2 (2  x) .
Show the coordinates of the points where the graph cuts the coordinate
axes.
[3]
(ii) Hence sketch the graph of y = f(2x), on a separate diagram, showing the
coordinates of the points where the graph cuts the coordinate axes.
[2]
3.
(i) Sketch the graph y 
1
.
x
[2]
1
on a separate diagram. Show the
x2
coordinates of any points where the graph cuts the coordinate axes.
(ii) Hence sketch the graph of y 
4.
[3]
(iii) Write down the equations of the asymptotes of the graph in (ii).
[2]
(i) On the same axes sketch the graphs of y  x( x  2)( x  5) and
y  x(1  x)( x  3) .
[6]
(ii) Use algebra to find the coordinates of the points where the graphs intersect.
[5]
5.
The diagram below shows the graph y = f(x), which has a turning point at (-2, 1)
and crosses the y-axis at (0, 5).
5
(-2, 1)
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EdExcel C1 Curves Assessment solutions
Sketch, on separate diagrams, each of the following graphs, showing the
coordinates of the turning point and the point at which the graph crosses the
y-axis in each case.
(i) y = f(x – 2)
[3]
(ii) y = 3f(x)
[3]
(iii) y = f( 12 x)
[3]
(iv) y = f(x) + 1
[3]
Total 40 marks
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EdExcel C1 Curves Assessment solutions
Sketching curves
Solutions to chapter Assessment
1. (i) y = x³
(ii) y = (x + 1)³
The graph of y = x³ is translated 1 unit to the left.
-1
(iii) y = x³ – 2
The graph of y = x³ is translated 2 units downwards.
-2
2. (i)
y  ( x  1)2(2  x )
When x = 0, y = 2
When y = 0, x = -1 (repeated) or 2.
When x is large and positive, y is large and negative.
When x is large and negative, y is large and positive.
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EdExcel C1 Curves Assessment solutions
y = f(x)
2
-1
2
(ii) The graph of y = f(x) is stretched in the x direction, scale factor
1
2
.
y = f(2x)
2
 21
3. (i)
y
1
1
x
1
1
, the graph of y 
is translated 2 units to the left.
x 2
x
When x = 0, y  21 .
(ii) To obtain y 
1
2
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EdExcel C1 Curves Assessment solutions
(iii) The asymptotes are x = -2 and y = 0.
4. (i)
y  x( x  2)( x  5 )
When x = 0, y = 0
When y = 0, x = 0, -2 or -5
When x is large and positive, y is large and positive.
When x is large and negative, y is large and negative.
y  x(1  x )( x  3)
When x = 0, y = 0
When y = 0, x = 0, 1 or -3
When x is large and positive, y is large and negative.
When x is large and negative, y is large and positive.
y  x(1  x )( x  3)
-5
y  x( x  2)( x  5 )
-3
-2
1
(ii) At intersections, x( x  2)( x  5 )  x(1  x )( x  3)
x( x 2  7 x  10)  x(3  2 x  x 2 )
x 3  7 x 2  10 x  3 x  2 x 2  x 3
2 x 3  9x 2  7 x  0
x(2 x 2  9 x  7)  0
x(2 x  7)( x  1)  0
x  0,  72 ,  1
When x = 0, y = 0
When x   72 , y   72   23  23 
When x = -1, y  1  1  4  4
63
8
The points of intersection are (0, 0),   72 , 63
8  and (-1, -4).
5. (i) y = f(x – 2)
The graph of y = f(x) is translated 2 units to the right.
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EdExcel C1 Curves Assessment solutions
y = f(x – 2)
(0, 1)
(ii) y = 3f(x)
The graph of y = f(x) is stretched in the y direction, scale factor 3.
y = 3f(x )
(0, 15)
(-2, 3)
(iii) y = f( 21 x)
The graph of y = f(x) is stretched in the x direction, scale factor 2.
y = f( 21 x )
(0, 5)
(-4, 1)
(iv) y = f(x) + 1
The graph of y = f(x) is translated 1 unit upwards.
y = f(x ) + 1
(0, 6)
(-2, 2)
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