Question 173: how do I find a parabola that passes through the points: (2,-3),(-2,9), and (1,3/2)?
Answer by AnlytcPhil(1806) (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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