Question 176899: Please help with this one...the directions are to find the vertex, the line of symmetry, the minimum and maximum value of the quadratic equation and then graph it. I know that once I get the vertex, I can find the rest. I have solved it close to the end, but I am not sure where to go from here...
f(x)=-2x^2+2x+7
I used the equation -b+- the square root of b^2-4ac divided by 2a
I got to -2+-the square root of 60 divided by 4. Now I am stuck...thank you for the help.
Found 2 solutions by scott8148, Earlsdon:
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! the equation you used is the "Quadratic Formula" __ it is used to find the roots (zeroes) of a quadratic equation
__ these will help with the graphing
the formula for the axis of symmetry is x=-b/(2a) __ this will give you the x-value of the vertex
__ substitute the x-value into the function to find the y-value
since the "a" coefficient is negative, the y-value of the vertex is the maximum value
the minimum value is -∞
Answer by Earlsdon(6294)  (Show Source):
You can put this solution on YOUR website! Find the vertex, line-of-symmetry, minimum or (not and) maximum of:
First, since your are not asked to find the x-intercepts, you didn't need to use the quadratic formula 
Notice that the coefficient of the  term is negative (-2), so this means that the parabola opens downward and, cosequently, the parabola will have a maximum, not a minimum.
Note, a parabola cannot have both a maximum and a minimum.
The maximum point will occur at the vertex of the parabola, so let's find the location of the vertex.
The x-coordinate of the vertex is given by: and, in this equation, a = -2, b = 2, so...
 Simplifying, we get:
 or  Now to find the y-coordinate, substitute this into the given equation, after replacing f(x) with y:
 Simplify this.
The vertex location is (0.5, 7.5) and this is also the location of the maximum point of the parabola.
The equation of the line of symmetry is simply the equation of the vertical line that passes through the vertex, or
Now let's look at the graph:
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