Asymptotes and Curve Sketching Past Paper Questions from AQA FP1

QUESTION 1 (2007 JUNE AQA FP1)

Vertical Asymptote when denominator is zero Horizontal Asymptote investigate the limit as x goes to infinity Horizontal Asymptote at y=3 Vertical Asymptote at x=-2

Curve passes through the point We now have the main features to sketch the graph

THIS IS THE RESULTING SKETCH

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

THIS IS THE RESULTING SKETCH

QUESTION 3 (2006 JUNE AQA FP1 Part of Question)

Curve passes through the x axis at the points THE GRAPH WILL CROSS THE X AXIS and

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)

Vertical Asymptote occurs when denominator is zero 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 (2005 JUNE AQA FP1)

Horizontal Asymptote is PARALLEL TO THE x AXIS. Investigate the limit as x goes to infinity Horizontal asymptote y=1 Dividing top and bottom by x squared

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!)