Question 468755
Find the vertex, the line of symmetry, the maximum or minimum value of the quadratic function and graph the function. 
f(x)=-2x^2+2x+4
f(x)= 1-x^2
...
Standard form of parabola: y=ħA(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex.
If the lead coefficient is>0, the parabola opens upwards and there is a minimum. If the lead coefficient is <0, the parabola opens downwards and there is a maximum. A is a multiplier which affects the steepness of the parabola.
..
f(x)=-2x^2+2x+4
completing the square
f(x)=-2(x^2-x+1/4)+4+1/2
f(x)=-2(x-1/2)^2+9/2
This is a parabola which opens downwards with vertex at (1/2,9/2). (maximum=9/2)
Axis of symmetry: x=1/2
..
f(x)= 1-x^2
f(x)=-x^2+1
Already in standard form,This is a parabola which opens downwards with vertex at (0,1). (maximum=1)
Axis of symmetry: x=0 or y-axis
See graph below as a visual check on answers
..

{{{ graph( 300, 300, -5, 5, -5, 5,-2x^2+2x+4,-x^2+1) }}}