SOLUTION: Can you help me find the equation of parabola axis vertical phases through (-1,0), (5,0), (1,8)?

Click here to see ALL problems on Quadratic-relations-and-conic-sections

Question 971171: Can you help me find the equation of parabola axis vertical
phases through (-1,0), (5,0), (1,8)?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39620) (Show Source):
You can put this solution on YOUR website!
Each point can give a system of general form quadratic equations.

Recognize that two of the given points are roots, before going on with the system of equations.
See what you might do with the given roots.

basic standard form format.
Using the two given roots,

Also recognize that the value for h will be in the MIDDLE of -1 and 5,
which is .
-
, based on standard form and one of the given points.

.

Again use the standard form system with knowledge of h=2.

.
Which is just the single equation.

Another a & k equation was found using (1,8); so put together this system to solve:

Solving for a and k should be easy.
.
Use this in either equation of the system to find k.

Now, a, h, k are known. The standard form equation can be written:

Answer by MathTherapy(10552) (Show Source):
You can put this solution on YOUR website!

Can you help me find the equation of parabola axis vertical
phases through (-1,0), (5,0), (1,8)?

General form of quadratic equation:
------- Using (- 1, 0) in general quadratic equation
------ eq (i)
------- Using (5, 0) in general quadratic equation
------ eq (ii)
------- Using (1, 8) in general quadratic equation
------ eq (iii)
- 2b = - 8 --------- Subtracting eq (iii) from eq (i)
, or b = 4
------- Substituting 4 for b in eq (i)
a = 4 - c ------- eq (iv)
25(4 - c) + 5(4) + c = 0 ------ Substituting 4 - c for a, and 4 for b in eq (ii)
100 - 25c + 20 + c = 0
- 25c + c + 120 = 0
- 24c = 0 - 120
- 24c = - 120
, or c = 5
a = 4 - 5 ------- Substituting 5 for c in eq (iv)
a = - 1
Therefore,
a = - 1
b = 4
c = 5
General quadratic equation: now becomes: