Question 764034
A quadratic equation has a parabola that has the roots x = 3 and x = -5.
 It passes through the point (1, -12).
 Determine it's equation.
:
Using the form ax^2 + bx + c = y, write equation for each pair
x=3; y = 0: 9a + 3b + c = 0
x=-.5; y=0: .25a - .5b + c = 0
x=1; y=-12: a + b + c = -12
:
Subtract the 3rd equation from the 1st equation
9a + 3b + c = 0
 a + b + c = -12
-----------------
8a + 2b = 12
simplify, divide by 2
4a + b = 6
:
Subtract the 3rd equation from the 2nd equation
.25a - .5b + c = 0
 a + b + c = -12
----------------
-.75a - 1.5b = 12
Simplify, divide both sides by -.75
a + 2b = -16
:
Two 2 unknown equations we can solve with elimination
Multiply the 1st by 2, subtract the 2nd
8a + 2b = 12
 a + 2b = -16
-----------------subtraction eliminates b, find a
7a = 28
a = 28/7
a = 4
:
Find b using a + 2b = -16
4 + 2b = -16
2b = -20
b = -10
:
Find c
4 - 10 + c = -12
c = -12 + 6
c = -6
:
Our equation: y = 4x^2 - 10x - 6
:
Graphically
{{{ graph( 300, 200, -5, 5, -20, 10, 4x^2-10x- 6)}}}