Question 173
<pre>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:
{{{ graph( 100, 100, -9, 3, -9, 10, -0.5*x^2-3*x+5)}}} 

Edwin