Question

Find a quadratic model for the set of values.

x
-2, 0,4

f(x)
1, -3 , 85

Answer 1

Okay well we know it is going to be in the form:\[f(x)=ax^2+bx+c\]Right?
Now each point gives us a new equation:\[1=a(-2)^2+b(-2)+c\]\[-3=a(0)^2+b(0)+c\]\[85=a(4)^2+b(4)+c\]So we have a system of equations... 3 equation and 3 unknowns. Can you solve for \(a\), \(b\), and \(c\)? Do you need help solving this system?

Answer 2

help please! :(

Answer 3

Okay.... So the second equation looks most easy to start with so:\[-3=a(0)^2+b(0)+c\]\[0 -3=+ 0 + c\]\[-3=c\]Now we have \(c\).

Answer 4

We can now make our system a bit simpler:\[1=a(-2)^2+b(-2)-3\]\[85=a(4)^2+b(4)-3\]

Answer 5

Do you understand what I've done so far?

Answer 6

somewhat I do..

Answer 7

What course are you taking right now?

Answer 8

algebra2

Answer 9

Have you ever had to solve a system of equations... like:
\[x+y=4\]\[4x-y=1\]

Answer 10

yes

Answer 11

Well what I gave you earlier is just a system of equations, but the variables are a, b, and c!

Answer 12

You use the same methods to solve for a, b, and c.

Answer 13

wait im really confused

Answer 14

okay.... let's just pretend that we're using x, y and z to make things simpler...

Answer 15

\[1=x(-2)^2+y(-2)+z\]\[-3=x(0)^2+y(0)+z\]\[85=x(4)^2+y(4)+z\]
So first let's get rid of the exponents and parenthesis.\[1=4x-2y+z\]\[-3=z\]\[85=16x+4y+z\]Can you solve this system of equations?