Write an equation for a parabola with​ x-intercepts (-3,0) and (2,0) which passes through the point (1,-16).

Answer 1

The equation has the form y = Ax² + Bx + C

Since (-3,0), (2,0), (1,-16) are points on the graph, we get the following system of equations:

9A - 3B + C = 0

4A + 2B + C = 0

A + B + C = -16

Subtracting the first 2 equations and subtracting the 3rd equation from the first, we obtain:

5A - 5B = 0

8A - 4B = 16

Since 5A - 5B = 0, 5A = 5B. So, A = B.

Therefore, 8A - 4A = 16.

4A = 16 So, A = B = 4

Since A + B + C = -16 and A = B = 4, we have 8 + C = -16. So, C = -24.

An equation of the parabola is y = 4x² + 4x - 24