Asymptotes and Curve Sketching Past Paper Qus

7/27/2019 Asymptotes and Curve Sketching Past Paper Qus

1/19

Asymptotes and Curve

Sketching

Past Paper Questions

from AQA FP1


2/19

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


4/19

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


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THIS IS THE

RESULTING

SKETCH


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QUESTION 2

(2007 JAN AQA FP1)


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2 1

xy

x


2

2

1 0

1

1

x

x

x

TWO vertical asymptotes

x=1 and x=-1


2

1

2 11 1x

x

xyx

as0

x

y

Horizontal asymptote y=0


8/19

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


( 2) 0

0

2

x x

x

x

TWO vertical asymptotes

x=0 and x=2


as

(1)(1)

(1)(1)

x

y

Horizontalasymptote

y=1

31

2(1 )(1 )

1(1 )

x x

x

y


12/19

( 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


as

6

1

x

y

Horizontal asymptote

is the line y=61

61

x

y


15/19

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


18/19

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.


19/19

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

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Asymptotes and Curve Sketching Past Paper Qus