Question 1080078

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 point on the graph then values are, x=-1 and y=5 that must satisfy y=ax^2+bx+c
Equation one being {{{ a(-1)^2+b(-1)+c=5 }}}

 (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 {{{ a(3)^2+b(3)+c=8 }}}

 (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 {{{ a(0)^2+b(0)+c=2 }}}

that gives the system of 3 equation which are
ONE {{{ a-b+c=5 }}}
TWO {{{ 9a+3b+c=8}}}
THREE {{{ c=2 }}}

then you start with ONE and THREE
{{{ a^2-b+c=5 }}}
{{{ c=2 }}} < 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!
<pre>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.