Math
posted by Allison A .
find the vertex line of symmetry the minimum naximum value of the quadratic function and graph the function f(x)=2x^2+2x+2.

A parabola has either a minimum or a maximum, but not both.
The axis of symmetry, which contains the vertex, is the line
x = b/2a = 2/(4) = 1/2
You should now be able to get the max/min and thus the vertex 
Express the function in canonical form by completing squares:
if f(x)=a(xh)²+k,
then
a>0 => the parabola is concave up, hence a minimum exists
a<0 => concave down, hence a maximum.
The location of maximum/minimum is given by the point (h,k).
The line of symmetry is x=h.
To proceed with completing the squares, extract and factor out the coefficient of x²:
f(x)=2(x²x) + 2
=2[(x1/2)²(1/2)²]+2
=2[(x1/2)²]+5/2
So the curve is concave down, h=1/2, k=5/2 and (h,k)=(1/2,5/2) is a maximum.
The line of symmetry is x=1/2. 
How do I find the Y coordinate and the x coordinates
is the vertix 1/2, 5/2? 
By completing squares, we get the two parameters h and k which represent the x and ycoordinates of the vertex.
In this case, they are (1/2,5/2), as you can see from:
f(x) = 2[(x1/2)²]+5/2