It's basically solve for three variables in a three equation system problem.
Plug point one into f(x) and you get:
3=a(1)^2+1b+c
Simplify and stick the constant on the right.
a+b+c=3
Point two:
8=a(2)^2+2b+c
again, simplify
4a+2b+c=8
Third:
12=a(-2)^2-2b+c
is
4a-2b+c=12
Now, you have options. Solve by Cramer's rule, substitution, or elimination. The system ends up being:
a + b + c = 3
4a + 2b + c = 8
4a - 2b + c =12(hopefully that stacks neatly in your email)
I'd use elimination, as it's much less messy. If you're interested in the steps for that, E-Mail me off list.
Cheers,
Alex
On Sep 12, 2007, at 11:49 AM, silverbabe_deluxe wrote:
hey..
how do I find a,b and c in
f(x)=ax^2+bx+c when I know the graph pass through point (1,3) (2,8)
and (-2,12)
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Alexander Cutshall