SOLUTION: The general equation os a quadratic polynomial is p(x)=ax(squared)+bx+c with a,b,c all real numbers and a doesn't = 0. In this problem we show that three points in the plane unique

Question 28313: The general equation os a quadratic polynomial is p(x)=ax(squared)+bx+c with a,b,c all real numbers and a doesn't = 0. In this problem we show that three points in the plane uniquely determine a quad polynomial, )just as two points in aplane uniquely determine a line).
Step 1. Suppose that we were given the three point (-1,1),(1,5), and (2,10). We want to find the quad poly P(x) which passes through these points. So, evaluate p at each x value. For example, since p(-1)must be 1 (why?), we get
p(-1)=a(-1) squared+b(-1)+c=1 (arrow) a-b+c=1.
Evaluating p(1) and p(2), we get two more equations. What are they?
p(1)=
p(2)=
Step 2: We now have _________________ linear equations in the _______________ unknowns___,__________, and _______. Thus, we can try to solve for _,_____, and ____. Once we have values for ___,______, and __, our polynomials will be determined.
Step 3: set up the augmented coefficient matrix. We can start with
M= [1 -1 1 1]
[? ? ? ?]
[? ? ? 10].
There are 6 values missing. Where do they come from? What are they? What variable does the first column represent? The second? The third?
Step 4:
Using elementary row operations, we transform M into the matrix M' that looks like
M'= [1 0 0 1]
[0 1 0 2]
[0 0 1 2].
I need to be able to explicity show the computations -- i.e the row of operations - leading from M tp M'.
From this it follows that a=____, b____, and c=___. Thus our polynomial p(x) is p(x)=?
Step 5:
Check the answer by evaluating p(x) with the a, b and c values determined in step 4. We should have p(-1)=1, p(1)=5, and p(2)=10.
Answer by longjonsilver(2297) (Show Source):
You can put this solution on YOUR website!
you have 3 points that are on a quadratic curve, (-1,1),(1,5), and (2,10).

If you put x=-1 into the quadratic, the answer for y is 1...this is the coordinate (-1,1). Similarly when x=1, then y=5 and also when x=2, then y=10.

We do not know the quadratic, so we are going build it up from first principles...

the quadratic is of the form

so using (-1,1) gives a-b+c=1
and using (1,5) gives a+b+c=5
and using (2,10) gives 4a+2b+c=10

these are your three "p"'s.

--> got 3 polynomials in 3 unknowns --> a,b,c.

matrix is

the columns are the coefficients of a,b and c respectively and the y-value.
the rows are the 3 equations.

OK, now you use Gaussian Elimination (GE) to get the matrix M'...look into your books for this. I hate GE - i spent years doing it at uni.

I have done GE for you... you struggle with it. It is the only way!
c=2
b=2
a=1

and i have checked the quadratic with the 3 sets of coordinates...they match.

jon.