Question 517710: Quadratic Function f(x)=-2x^2+2x+6
The x-coordinate of the vertex is?
The y-coordinate of the vertex is?
The equation of the line of symmetry is?
The maximum/minimum of f(x) is?
The value, f(1/2)=13/2 is?
Is it a maximum or a minimum?
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! Quadratic Function f(x)=-2x^2+2x+6
The x-coordinate of the vertex is?
x = -b/(2a)
x = -2/(2(-2))
x = -2/(-4)
x = 1/2
.
The y-coordinate of the vertex is?
set x=1/2 and solve for y:
y=-2x^2+2x+6
y=-2(1/2)^2+2(1/2)+6
y=-2(1/4)+1+6
y=-1/2+7
y=-1/2+14/2
y=13/2
.
The equation of the line of symmetry is?
x = 1/2
The maximum/minimum of f(x) is?
6.5
.
The value, f(1/2)=13/2 is?
Is it a maximum or a minimum?
By inspecting the original equation:
f(x)=-2x^2+2x+6
Since the coefficient associated with the x^2 is NEGATIVE
we now know it is a parabola that opens downwards.
therefore 13/2 is the MAXIMUM
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