SOLUTION: How do you find real numbers a,b and c so that the graph of the function {{{ y=ax^2+bx+c }}} contains the points (-1,5), (3,8) and (0,2)? My process went as (-1,5) is a poin

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Question 1080078: How do you find real numbers a,b and c so that the graph of the function contains the points (-1,5), (3,8) and (0,2)?
My process went as
(-1,5) is a point on the graph then values are, x=-1 and y=5 that must satisfy y=ax^2+bx+c
Equation one being
(3,8) is a point on the graph then values are, x=3 and y=8 that must satisfy y=ax^2+bx+c
Equation two being
(0,2) is a point on the graph then values are, x=0 and y=2 that must satisfy y=ax^2+bx+c
Equation three being
that gives the system of 3 equation which are
ONE
TWO
THREE
then you start with ONE and THREE

< multiply each side by -1
and add to get a-b=3
Im not sure if thats the way to go where you then use equation 3 to eliminate c from equation 2 or if there's another process to get solutions of a, b and c? I'm thankful for any advice on this problem!
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39620)   (Show Source): You can put this solution on YOUR website!
First use the point of one of the given roots, (0,2).




-
Make two equations with each of the other given points.


-


-

, simplifiable,

-
Now solve this system:
.
.
.
Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!

How do you find real numbers a,b and c so that the graph of the function contains the points (-1,5), (3,8) and (0,2)?
My process went as
(-1,5) is a point on the graph then values are, x=-1 and y=5 that must satisfy y=ax^2+bx+c
Equation one being
(3,8) is a point on the graph then values are, x=3 and y=8 that must satisfy y=ax^2+bx+c
Equation two being
(0,2) is a point on the graph then values are, x=0 and y=2 that must satisfy y=ax^2+bx+c
Equation three being
that gives the system of 3 equation which are
ONE
TWO
THREE
then you start with ONE and THREE

< multiply each side by -1
and add to get a-b=3
Im not sure if thats the way to go where you then use equation 3 to eliminate c from equation 2 or if there's another process to get solutions of a, b and c? I'm thankful for any advice on this problem!
Great job in getting up to that point.

a - b + c = 5 ----- eq (i)

9a + 3b + c = 8 ----- eq (ii)

c = 2 ------ eq (iii)

Now, just substitute 2 for c in eq (i) to get: a – b = 3 ------- eq (iv)

Then, substitute 2 for c in eq (ii) to get: 9a + 3b = 6, which reduces to 3a + b = 2 ------- eq (v)



Now, ADD eqs (v) & (iv) to ELIMINATE b

From there, you should be able to find the values of a and b, you already have the value of c, so you have everything to form the equation. 

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