Question 352607: Find the vertex, the line of symmetry, the maxiumum or minimum value of the quadratic function, and graph the function.
f(x)=-2x^2+2x+3
The x-coordinate of the vertex is:
The y-coordinate of the vertex is:
The equation of the line of symmetry is x=
The maximum/minimum of f(x) is:
The value, f(1/2)=7/2 is minimum or maximum?
So far, I have tried to work the problem. This is what I have:
f(x)=-2x^2+2x+3, so you use (-b/2a,4ac-b^2/4a)
Therefore, if -b/2a=-(2)/2(-1)=1
Then put 1 back into the equation.
f(1)=-2(1)^2+2(1)+3
Using FOIL, I come up with 1.
This where I get confused. Is the x-coordinate 1? How do i come up with y-coordinate?
Is the line of symmetry also 1?
I think I confuse myself the most when I try to apply the FOIL method?
Any help is so very much appreciated! Thanks in advance!
Found 2 solutions by Fombitz, stanbon: Answer by Fombitz(32388) (Show Source): Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Find the vertex, the line of symmetry, the maxiumum or minimum value of the quadratic function, and graph the function.
f(x)=-2x^2+2x+3
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vertex occurs at x = -b/(2a) = -2/(2*-2) = 1/4
This is a maximum
The x-coordinate of the vertex is:1/4
The y-coordinate of the vertex is:f(1/4) = 3.375
The equation of the line of symmetry is x= 1/4
The maximum/minimum of f(x) is: max = 3.375
The value, f(1/4)=3.375 is maximum because the coefficient of x^2 is negative.
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Cheers,
Stan H.
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