1 / 19

# Asymptotes and Curve Sketching Past Paper Questions from AQA FP1 - PowerPoint PPT Presentation

Asymptotes and Curve Sketching Past 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.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' Asymptotes and Curve Sketching Past Paper Questions from AQA FP1' - valentine-burris

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

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

QUESTION 1 (2007 JUNE AQA FP1)

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

We now have the main features to sketch the graph

QUESTION 2 (2007 JAN AQA FP1)

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)

Curve passes through the x axis at the points

and

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

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)

Investigate the limit as x goes to infinity

Dividing top and bottom by x squared

Horizontal asymptote y=1

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!)