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Asymptotes and Curve Sketching Past Paper Questions from AQA FP1PowerPoint Presentation

Asymptotes and Curve Sketching Past Paper Questions from AQA FP1

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### Asymptotes and Curve SketchingPast Paper Questions from AQA FP1

QUESTION 1 (2007 JUNE AQA FP1)

Vertical Asymptote when denominator is zero

Vertical Asymptote at x=-2

Horizontal Asymptote investigate the limit as x goes to infinity

Horizontal Asymptote at y=3

Curve passes through the point

Curve passes through the point

We now have the main features to sketch the graph

QUESTION 2 (2007 JAN AQA FP1)

Vertical Asymptote when denominator is zero

TWO vertical asymptotes

x=1 and x=-1

Horizontal Asymptote investigate the limit as x goes to infinity

Horizontal asymptote y=0

(2006 JUNE AQA FP1 Part of Question)

Vertical Asymptote when denominator is zero

TWO vertical asymptotes

x=0 and x=2

Horizontal Asymptote investigate the limit as x goes to infinity

Horizontal asymptote

y=1

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)

QUESTION 4 (2006 JAN AQA FP1) the sketch that would be obtained.

Vertical Asymptote occurs when denominator is zero the sketch that would be obtained.

A vertical asymptote is the line x=1

Horizontal Asymptote investigate the limit as x goes to infinity

Horizontal asymptote is the line y=6

When x=0 we see that y=0 also. The graph passes through (0,0)

If we attempt to find stationary values we find there are none! The gradient is always negative. (try it!)

The function is monotonic Decreasing.

QUESTION 5 (0,0)(2005 JUNE AQA FP1)

Horizontal Asymptote is PARALLEL TO THE x AXIS. (0,0)

Investigate the limit as x goes to infinity

Dividing top and bottom by x squared

Horizontal asymptote y=1

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 because

Suggests that

But this equation does not

have REAL roots

Concluding, The denominator can not be zero and so there

are no vertical asymptotes.

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 WILL cross the horizontal asymptote

We would find the values

of x when y=0 (check them!)

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