Maths Coursework: Curve Fitting Essay Sample
- Pages: 4
- Word count: 1,099
- Rewriting Possibility: 99% (excellent)
- Category: mathematics
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Introduction of TOPIC
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:
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
I worked the equation out to be Y=x2-8x+15. I then plotted this graph using omnigraph as below:
I then went onto the second task of finding the equation of the quadratic graph which passes through (0,9) and touches the x-axis at (3,0). I used the same method as before, of drawing what I thought the graph would look like:
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:
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:
n put the details I knew from the graph into three separate equations. I then labelled them a, b and
To work out these equations I had to get rid of the c’s, so I made two new equations by working out the equation c-a, and then a-b, to create equation d and e.
I then had to get either the b’s or the a’s to an equal value, so that I could work out the remaining letter. I did this by multiplying equation e by 2, so that both equations had 6b’s. This new equation was labelled f. By working out equation f and d I could work out that 18=18a as below:
I substituted a=1 into e to find:
I then substituted a=1 and b=-5 into a so that I could work out c.
So a=1, b=5 and c=4. This means that on the specified graph the equation is :
1. Firstly use the equation Y=ax2+bx+c
2. Substitute the three separate co-ordinates into three separate equations of the one used in point 1 above. For instance the co-ordinates are given in the form (x,b), x is the point that the graph crosses the x axis and b is the point where the graph crosses the y axis. So the points can be substituted so the equation is: B=ax2+bx+c
3. Number these three equations 1,2,3. You know need to get rid of the c’s, so two new equations can be made by subtracting equation 1 from 3, and labelling this equation as 4, and then subtracting equation 2 from 1, and labelling this equation as 5.
4. Now it is important to get any one letter to have an equivalent value, so use b values. Multiply equation 5 by the number of b’s there are in equation 4 and label this new equation 6. Then multiply equation 4 by the number of b’s there are in equation 5 and label this equation 7.
5. You now have to get rid of the b’s from both equations to leave you with the value of a. You do this by either adding or subtracting equation 6 and 7 together.
6. Now you have the value of -a. To find the value of a alone, divide the number value you have by -a.
7. Substitute the value for a into equation 4. From this work out b using the above rules.
8. Now substitute the values of a and b into equation 1 and then work out the value of c.
9. Now you have the values of a, b and c. Substitute these values into the equation: Y=ax2+bx+c, and this is the equation of the graph.