Question

how do I find a quadratic model for a set of values?
(-1,1) , (1,1) (3,9)

Answer 1

The general quadratic equation is

y = a*x^2 + b*x + c

We'll use this to set up our equations

Answer 2

The first point is (-1,1). So x = -1 and y = 1


y = a*x^2 + b*x + c


y = a*(-1)^2 + b*(-1) + c ... plug in x = -1


y = a*(1) + b*(-1) + c


y = 1a - 1b + c


1 = 1a - 1b + c ... plug in y = 1


1 = a - b + c


a - b + c = 1


So the first equation is `a - b + c = 1`

Answer 3

The second point is (1,1). So x = 1 and y = 1


y = a*x^2 + b*x + c


y = a*(1)^2 + b*(1) + c ... plug in x = 1


y = a*(1) + b*(1) + c


y = 1a + 1b + c


1 = a + b + c ... plug in y = 1


a + b + c = 1


The second equation is `a + b + c = 1`

Answer 4

And finally, the third point is (3,9), so...


x = 3
y = 9



y = a*x^2 + b*x + c


y = a*(3)^2 + b*(3) + c ... replace x with 3


y = 9a + 3b + c


9 = 9a + 3b + c ... replace y with 9


9a + 3b + c = 9


The third equation is `9a + 3b + c = 9`

Answer 5

we now have this system of equations


\[\Large \begin{cases}a-b+c = 1\\a+b+c = 1\\9a+3b+c = 9\end{cases}\]

hopefully you're able to solve for a,b,c ?

Answer 6

yep! thanks alot :D

Answer 7

you're welcome