SOLUTION: Find the vertex, the line of symmetry, the maximum or minimum calue of the quadratic function, and graph the function. f(x)=-2x^2+2x+2

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the vertex, the line of symmetry, the maximum or minimum calue of the quadratic function, and graph the function. f(x)=-2x^2+2x+2      Log On


   



Question 560874: Find the vertex, the line of symmetry, the maximum or minimum calue of the quadratic function, and graph the function.
f(x)=-2x^2+2x+2

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
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+2
complete the square
y=-2(x^2-x+1/4)+2+1/2
y=-2(x-1/2)^2+5/2
This is an equation for a parabola of the standard form: y=-A(x-h)^+k, (hk) being the (x,y) coordinates of the vertex. Negative lead coefficient means parabola opens downwards.
For given function:
vertex: (1/2,5/2)
axis (line) of symmetry: x=1/2
maximum value: 5/2
see graph below:
+graph%28+300%2C+300%2C+-4%2C+4%2C+-4%2C+4%2C-2%28x-1%2F2%29%5E2%2B5%2F2%29+