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:
(0,9)
(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
-2=a(22)+b(2)+c
-2=4a+2b+c
c
10=a(-12)+b(10)+c
10=1a-1b+c
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.
d
6=24a-6b
e
6=21a+3b
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:
f
12=42a+6b
f+d
18=18a
a=1
I substituted a=1 into e to find:
6=21+3b
Therefore
6-21=3b
-15=3b
b=-5
I then substituted a=1 and b=-5 into a so that I could work out c.
10=1+5+c
Therefore
10=6+c
10-6=4
c=4
So a=1, b=5 and c=4. This means that on the specified graph the equation is :
Y=x2-5x+4
Method
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Firstly use the equation Y=ax2+bx+c
-
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
- 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.
- 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.
- 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.
- Now you have the value of –a. To find the value of a alone, divide the number value you have by –a.
- Substitute the value for a into equation 4. From this work out b using the above rules.
- Now substitute the values of a and b into equation 1 and then work out the value of c.
-
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.
