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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