# Maths Coursework: Curve Fitting

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Introduction

## Maths Coursework: Curve Fitting

I was firstly given the task of finding the equation of the quadratic graph which passes through the points (5,0), (3,0), (0,15). To solve this I began by drawing a rough sketch of what I thought the graph would look like with these points, as below:

(0,15)

(3,0) (5,0)

I worked out that the graph would look like this, and next I worked out the formula by putting the numbers I knew into brackets, and then expanding them as below. I did this because by looking at the graph you can see that when Y=0, X=5 or X=3.

Y = (x-5)(x-3) = 0

- x2-3x-5x+15
- =x2-8x+15

I worked the equation out to be Y=x2-8x+15. I then plotted this graph using omnigraph as below:

Middle

(3,0)

I then put the numbers into brackets again (as below), because I worked out that when Y=0, X=3, and no other number. Then once again expanded the brackets to find the formula:

Y=(x-3)(x-3) = 0

I worked out the formula to be:

Y=x2-6x+9

I could be sure that this was the correct equation because the co-ordinate was (0,9) which shows that the graph passes through +9, and the above equation proves this.

I decided to find the equation of the graph which passes through the points (-1,10), (2,-2), (5,4) before I worked out a method. I started by sketching what I thought the graph would look like. I also realised that the equation for all graphs is:

Y= ax2+bx+c

(-1,10)

(5,4)

(2,-2)

I then put the details I knew from the graph into three separate equations. I then labelled them a, b and c.

a

4=a(52)+b(5)+c

4=25a+5b+c

b

Conclusion

Page:

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

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