50g

I found it more fun to just pull out my 50g, enter a [3 8 12] vector 
for the right-side constants, use the MTRW to construct a 3X3 matrix 
for the left side constants, and then "divide" it into the vector to 
get [2 -1 3] as the values for A, B, and C.

Just for grins, I ran it through the linear system solver too and, 
amazingly, got the same answer :-)

Dick

--- In 50g@yahoogroups.com, Alexander Cutshall <alexc@...> wrote:
>
> 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)
> >
> 
> 
> ------
> "I love America more than any other country in this world, and,  
> exactly for this reason, I insist on the right to criticize her  
> perpetually."
> - James Baldwin
> 
> Alexander Cutshall
> alexc@...
>