SOLUTION: Could you explain to me how to construct the equation of a parabola that has x-axis intercepts at (-2,0) and (4,0) ?

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Could you explain to me how to construct the equation of a parabola that has x-axis intercepts at (-2,0) and (4,0) ?      Log On


   

Question 131738: Could you explain to me how to construct the equation of a parabola that has x-axis intercepts at (-2,0) and (4,0) ?
Found 2 solutions by solver91311, stanbon:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
If the parabola intercepts the axis at (-2,0), that means the value of the function at x = -2 is f(-2) = 0. Then one of the factors of the equation must be
x%2B2. Likewise, the other factor must be x-4, so the equation becomes:

y=f%28x%29=%28x%2B2%29%28x-4%29

y=x%5E2-2x-8

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Could you explain to me how to construct the equation of a parabola that has x-axis intercepts at (-2,0) and (4,0) ?
--------------
Rule: If "a" is a root, (x-a) is a factor.
--------------
You are given two roots: x = -2 and x=4
So you have two factors (x+2) and (x-4)
----------
EQUATION:
y = (x+2)(x-4)
y = x^2-2x-8
====================
graph%28400%2C300%2C-10%2C10%2C-10%2C10%2Cx%5E2-2x-8%29
====================
Cheers,
Stan H.