Question 173:  how do I find a parabola that passes through the points: (2,-3),(-2,9), and (1,3/2)? 
 Answer by AnlytcPhil(1810)      (Show Source): 
You can  put this solution on YOUR website! How do I find a parabola that passes through the points: 
(2,-3),(-2,9), and (1,3/2)? I can not find any examples in my book. 
Start with the general equation of a parabola: 
                  y = Ax² + Bx + C  
Substitute in the point (2,-3), that is, substitute 2 for x and -3 for y 
                 -3 = A(2)² + B(2) + C 
                 -3 = A(4) + 2B + C 
                 -3 = 4A + 2B + C 
(1)     4A + 2B + C = -3 
 Substitute in the point (-2,9), that is, substitute -2 for x and 9 for y 
                  y = Ax² + Bx + C 
                  9 = A(-2)² + B(-2) + C 
                  9 = A(4) - 2B + C 
                  9 = 4A - 2B + C 
(2)     4A - 2B + C = 9 
Substitute in the point (1,3/2), that is, substitute -2 for x and 9 for y 
                  y = Ax² + Bx + C 
                3/2 = A(1)² + B(1) + C 
                3/2 = A(1) + B + C 
                3/2 = A + B + C 
(3)       A + B + C = 3/2 
So you have this system of equations: 
(1)     4A + 2B + C = -3 
(2)     4A - 2B + C = 9 
(3)       A + B + C = 3/2
 
Do you know how to solve this system? I will assume you do. 
If you don't, post again.
 
Answers:   A = -1/2, B = -3,  C = 5
 
So substitute these values into the general equation
 
                  y = Ax² + Bx + C 
and get 
                  y = -1/2x² - 3x + 5
 
Here's the graph: 
  
 
Edwin   
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