# Writing a quadratic function using a table of values

• Jul 25th 2010, 01:51 PM
wiseguy
Writing a quadratic function using a table of values
"A quadratic function f contains these points.
x y
3 36.7
5 27.5
7 20.7
9 16.3
11 14.3

Find the particular equation for f(x) as a function of x."

Here is my response:
ax^2+bx+c
36.7=a(3)^2+b(3)+c
36.7=9a+3b+c
36.7-9a-3b=c

27.5=a(5)^2+b(5)+(36.7-9a-3b)
-9.2=16a+2b
(-9.2-16a)/2=b

20.7=a(7)^2+((-9.2-16a)/2)7+(36.7-9a-3((-9.2-16a)/2))
20.7=49a+(-4.6-8a)7+(6.7-9a-3(-4.6-8a))
20.7=49a-32.2-56a+6.7-9a+13.8+24a
20.7=-11.7+8a
32.4=8a
a=4.05

16.3=4.05(9)^2+((-9.2-16(4.05))/2)(9)+(36.7-9(4.05)-3((-9.2-
16(4.05))/2))
16.3=4.05(9)^2-37(9)+111.25

Unfortunately, after double checking it into the calculator (using quadratic regression), the real equation is \$\displaystyle 0.3x^2+-7x+55\$ ...I'm at a bit of a loss.

Any help would be greatly appreciated :)
• Jul 25th 2010, 02:07 PM
pickslides
I also get \$\displaystyle y = 0.3x^2 - 7x + 55\$ using the least squares regression model. Do you know this model?
• Jul 25th 2010, 02:46 PM
wiseguy
No I do not, but I would love to find out...
• Jul 25th 2010, 03:05 PM
pickslides
Check this example. Curve Fitting, Part 1

Using technology will be easier though, I found your equation very easily using ms-excel.
• Jul 25th 2010, 04:20 PM
1005
I think you're given that those are points on a quadratic equation, ie this is not an estimation. just assume the form ax^2 + bx + c = y and make 3 equations using any 3 y/x relationships you know and solve the system of 3 unknowns for a, b, and c.

I ran an rref on a matrix like so and got the right answer:
[3^2, 3, 1, 36.7]
[5^2, 5, 1, 27.5]
[7^2, 7, 1, 20.7]
• Jul 25th 2010, 07:16 PM
wiseguy
36.7=a(3)^2+b(3)+c
27.5=a(5)^2+b(5)+c
20.7=a(7)^2+b(7)+c

36.7-9a-3b=27.5-25a-5b
-27.5+25a+5b=-20.7+49a+7b

36.7-9a-3b=-20.7+49a+7b
57.4-58a=10b
5.74-5.8a=b

36.7-9a-3(5.74-5.8a)=-20.7+49a+7(5.74-5.8a)
36.7-9a-17.22+17.4a=-20.7+49a+40.18-40.6a
80.36=-9a
a=-8.928888888

b=-46.04755555