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7/27/2019 Asymptotes and Curve Sketching Past Paper Qus
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Asymptotes and Curve
Sketching
Past Paper Questions
from AQA FP1
7/27/2019 Asymptotes and Curve Sketching Past Paper Qus
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QUESTION 1
(2007 JUNE AQA FP1)
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3 1
2
xy
x
Vertical Asymptote when denominator is zero
2 0
2
x
x
Horizontal Asymptote investigate the limit as x goes to infinity
as
3
1
x
y
Horizontal Asymptote at y=3
Vertical Asymptote at x=-2
1
2
33 1
2 1
x
x
xy
x
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when 0
3(0) 1 10 2 2
x
y
Curve passes through the point1
(0, )2
when y 03 1
02
0 3 1
1
3
x
x
x
x
Curve passes through the point1
( ,0 )
3
We now have the main features to sketch the graph
7/27/2019 Asymptotes and Curve Sketching Past Paper Qus
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THIS IS THE
RESULTING
SKETCH
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QUESTION 2
(2007 JAN AQA FP1)
7/27/2019 Asymptotes and Curve Sketching Past Paper Qus
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2 1
xy
x
Vertical Asymptote when denominator is zero
2
2
1 0
1
1
x
x
x
TWO vertical asymptotes
x=1 and x=-1
Horizontal Asymptote investigate the limit as x goes to infinity
2
1
2 11 1x
x
xyx
as0
x
y
Horizontal asymptote y=0
7/27/2019 Asymptotes and Curve Sketching Past Paper Qus
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2
1
xy
x
THIS IS THE
RESULTING
SKETCH
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QUESTION 3
(2006 JUNE AQA FP1 Part of Question)
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when y 0
( 1)( 3)0
( 2)
0 ( 1)( 3)
1 or 3
x x
x x
x x
x x
Curve passes through the x axis at the points
( 1,0)
THE GRAPH WILL CROSS THE X AXIS
and (3,0)
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( 1)( 3)
( 2)
x xy
x x
Vertical Asymptote when denominator is zero
( 2) 0
0
2
x x
x
x
TWO vertical asymptotes
x=0 and x=2
Horizontal Asymptote investigate the limit as x goes to infinity
as
(1)(1)
(1)(1)
x
y
Horizontalasymptote
y=1
31
2(1 )(1 )
1(1 )
x x
x
y
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( 1)( 3)
( 2)
x xy
x x
The graph was not required in this exam question but this is the sketch that
would be obtained.
We would need to differentiate and equate to zero in order to find that the
stationary point is a min at (1,4)
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QUESTION 4 (2006 JAN AQA FP1)
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6
( 1)
xy
x
Vertical Asymptote occurs when denominator is zero
1 0
1
x
x
A vertical asymptote is the line x=1
Horizontal Asymptote investigate the limit as x goes to infinity
as
6
1
x
y
Horizontal asymptote
is the line y=61
61
x
y
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6
( 1)
xy
x
When x=0 we seethat y=0 also. The
graph passes through
(0,0)
If we attempt to find stationary valueswe find there are none! The gradient is
always negative. (try it!)
The function is monotonic Decreasing.
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QUESTION 5
(2005 JUNE AQA FP1)
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2
2
4
9
x xy
x
Horizontal Asymptote is PARALLEL TO THE x AXIS.
Investigate the limit as x goes to infinity
as
1
1
x
y
Horizontal asymptote y=1
2
4
9
1
1
x
x
y
Dividing top and bottom by x squared
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Explain why the graph has no asymptote parallel
to the y axis.
An asymptote which is parallel to the y axis is vertical
and would occur when the denominator is zero. Is this
possible here?
No because2
2
4
9
x xy
x
Suggests that
2
9 0x But this equation does not
have REAL roots
Concluding, The denominator can not be zero and so there
are no vertical asymptotes.
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The graph was not required in this exam question but this is the sketch that
would be obtained.
In this graph when y=1 we find x=2.25 (check this!) So the graph WILLcross the horizontal asymptote
We would find the values
of x when y=0 (check them!)
2
2
4
9
x xy
x