Question 1008243
<pre>
You want to find the equation of this parabola. Scroll down:

{{{drawing(250,400,-5,5,-6,10,
locate(-4.1,8.3,"(-2,8)"),
locate(2.1,4.3,"(2,4)"),
locate(1.1,-4+.3,"(1,-4)"),
graph(250,400,-5,5,-6,10,3x^2-x-6),

circle(2,4,0.15),circle(2,4,0.13),circle(2,4,0.11),circle(2,4,0.09),circle(2,4,0.07),circle(2,4,0.05),circle(2,4,0.03),circle(2,4,0.01),

circle(-2,8,0.15),circle(-2,8,0.13),circle(-2,8,0.11),circle(-2,8,0.09),circle(-2,8,0.07),circle(-2,8,0.05),circle(-2,8,0.03),circle(-2,8,0.01),

circle(1,-4,0.15),circle(1,-4,0.13),circle(1,-4,0.11),circle(1,-4,0.09),circle(1,-4,0.07),circle(1,-4,0.05),circle(1,-4,0.03),circle(1,-4,0.01) )}}}

(2, 4),(-2, 8),and (1, -4)

{{{y}}}{{{""=""}}}{{{ax^2+bx+c}}}

For the point (2, 4), substitute x = 2 and y = 4

{{{4}}}{{{""=""}}}{{{a(2)^2+b(2)+c}}}

For the point (-2, 8), substitute x = -2 and y = 8

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

For the point (1, -4), substitute x = 1 and y = -4

{{{-4}}}{{{""=""}}}{{{a(1)^2+b(1)+c}}}

That gives us the system of three equations in three
unknowns a,b, and c

{{{system(4=a(2)^2+b(2)+c,8=a(-2)^2+b(-2)+c,-4=a(1)^2+b(1)+c)}}}

Simplify those equations:

{{{system(4=a(4)+b(2)+c,8=a(4)+b(-2)+c,-4=a(1)+b(1)+c)}}}

{{{system(4=4a+2b+c,8=4a-2b+c,-4=a+b+c)}}}

Swap the sides of the equations so the system will look "normal":

{{{system(4a+2b+c=4,4a-2b+c=8,a+b+c=-4)}}}

Solve that by elimination.  Then substitute the values you get
for a, b, and c in 

{{{y}}}{{{""=""}}}{{{ax^2+bx+c}}}

If you run into any trouble finishing, ask me in the thank-you note 
form below, and I'll get back to you by email.  

Edwin</pre>