SOLUTION: This is more of a question than math problem, but here goes anyway, If I have a quadratic equation (y = x^2 – 2x – 8)) that graphs to a parabola with a minimum value at the ver

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: This is more of a question than math problem, but here goes anyway, If I have a quadratic equation (y = x^2 – 2x – 8)) that graphs to a parabola with a minimum value at the ver      Log On


   

Question 163367: This is more of a question than math problem, but here goes anyway,
If I have a quadratic equation (y = x^2 – 2x – 8)) that graphs to a parabola with a minimum value at the vertex (1, -9), how would you find the maximum value for the equation?

Thanks for any help.
Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
if the quadratic equation has a minimum value, there is no maximum value.
general form of quadratic equation is a*x^2 + b*x + c.
if it's -a*x^2, then the parabola has a maximum value.
if it's +a*x^2, then the parabola has a minimum value.
the turning point is given by -b/2a.
your equation is y = x^2 - 2x - 8
a = 1
b = -2
c = -8
-b/2a = 2/2 = 1 which is the x value of the turning point.
solve for the y value in the equation.
y = 1^2 - 2*1 - 8
y = -9 as you showed.
a graph of this equation looks like this:
see below graph for further comments.
graph+%28800%2C800%2C-10%2C10%2C-10%2C10%2Cx%5E2-2%2Ax-8%29
as you can see, the graph has a minimum but no maximum.
if the equation were -x^2+2*x+8 then the graph would look like this:
scan below the graph for further comments.
graph%28800%2C800%2C-10%2C10%2C-10%2C10%2C-x%5E2%2B2%2Ax%2B8%29
the turning point in this case is -b/2a where
b = +2
2a = -2
turning point x value = -(2)/-2 = 1
y turning point = -(1)^2 +2*1)+8 = 9
turning point coordinate = (1,9)