SOLUTION: I am studying for a final in College Algebra, this question comes from my review sheet. The final is tomorrow so any immediate help is GREATLY appreciated. Find the equation of

I am studying for a final in College Algebra, this
question comes from my review sheet. The final is 
tomorrow so any immediate help is GREATLY appreciated.
 
Find the equation of the parabola which passes through 
the points (0,4),(-2,0),(-3,1). Write the equation in 
quadratic form.

The general equation of a parabola is

   y = Ax² + Bx + C

Substitute the point (x, y) = (0, 4) into this expression.
That is substitute 0 for x and 4 for y
   
   y = Ax² + Bx + C
   4 = A(0)² + B(0) + C
   4 = C

Substitute the point (x, y) = (-2, 0) into this expression.
That is substitute -2 for x and 0 for y
   
   y = Ax² + Bx + C
   0 = A(-2)² + B(-2) + C
   0 = A(4) - 2B + C
   0 = 4A - 2B + C

Substitute the point (x, y) = (-3, 1) into this expression.
That is substitute -3 for x and 1 for y
   
   y = Ax² + Bx + C
   1 = A(-3)² + B(-3) + C
   1 = A(9) - 3B + C
   1 = 9A - 3B + C

So we have this system of equations. 

   4 = C
   0 = 4A - 2B + C
   1 = 9A - 3B + C
 
or
 
   C = 4
   4A - 2B + C = 0
   9A - 3B + C = 1

Solve that and get A = 1, B = 4, C = 4

Substitute these in

   y = Ax² + Bx + C

   y = 1x² + 4x + 4

Erase the 1 coefficient on x²

   y = x² + 4x + 4

Edwin McCravy
AnlytcPhil@aol.com