50g

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)

 4a -  2b +  c =12

I think that this little program can help you... it`s from a
minichallenge realized in a latin american web page...

As arguments you have to put the three points in lists in the
following way:

7:
6:
5:
4:
3: {x1 y1}
2: {x2 y2}
1: {x3 y3}

It`s the source code...

\<< + + EVAL SWAP DUP
SQ 1 \->V3 4 ROLL DUP
SQ 1 \->V3 6 ROLL DUP
SQ 1 \->V3 SWAP ROT 3
ROW\-> INV 4 ROLLD
\->V3 * -3 FS?C 'Y'
ROT V\-> SWAP 'X^2' *
ROT 'X' * + SWAP +
= SWAP
\<< -3 SF
\>> IFT
\>>

Greetings from Venezuela...

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@...
>

To solve your problem, first put your calculator in RPN mode and soft
menu (flag 117 set).
first enter a vector with the values of y. (wherever you see a space
pres the [SPC] key and notice LS Left Shift and RS Right Shift.)

LS[]3 8 12 ENTER

enter a vector with X^2 X and 1 and duplicate it.
LS[] 'X^2' 'X' 1 ENTER ENTER

enter X=1 and substitute this value in the vector.

'X=1' ENTER
RS ALG NXT SUBST

swap the contents in level 1 and level 2, duplicate the vector in
level 1, enter X=2 and substitute this value in the vector.

Press the right arrow of the cursor key to swap.
ENTER
'X=2' ENTER
SUBST

swap one last time, enter X=-2 and substitute this value in the
vector.

Press the right arrow of the cursor key to swap.
'X=-2' ENTER
SUBST

At this point you have four vectors in the stack, the last three are
rows of the matrix for the system. Let's form that matrix.

3
RS MTH MATRX ROW ROW->

to solve this system just press the divided by key, here we represent
it by /

/

the result is the vector [2 -1 2]

a=2, b=-1 and c=2

you can check the solution with the function PEVAL (Polynomial
EVALuation) you need the vector of coeficients in level 2 in this
case [2 -1 2] and the value of X (like 1) in level 1 and PEVAL find
the value of f(x).