Question 395569
    find the vertex, the line of symmetry, and the maximim or minimum value of the quadratic function, and graph the function on paper.
    f(x) = 2x^2+2x+4
    what is the x-coordinate of the vertex? (type a simplified fraction)

Just by inspection of the equation, you should know that f(x) is a parbola that opens upwards because the coefficient of the x^2 term is positive, therefore, it has a minimum. Conversely, if the coefficient is negative, the parabola will open downward and therefore, have a maximum.

There are 2 methods for finding the coordinates of the vertex.
The first method uses the formula -b/2a, a, being the coefficient of the x^2 term, and,b, being the coefficient of the x term. This formula gives you the x coordinate of the vertex. Substitute this value into the equation to get the y coordinate.

The second method is to complete the square and put it into this form:
y=(x-h)^2+k,(h,k) being the (x,y) coordinates of the vertex.


Solve y=2x^2+2x+4 using the first method
a=2
b=2
-b/2a=-2/4=-1/2
x=-1/2
y=2(-1/2)^2+2(-1/2)+4
 =1/2-1+4=3+1/2
(x,y) coordinates of the vertex are (-1/2,3+1/2)
3+1/2 is the minimum value, and the line of symmetry is x=-1/2

Solve y=2x^2+2x+4 using the second method (completing the square)
first, factor out 2 from the x^2 and x terms, but leave the constant, 4, outside
2(x^2+x+  )+4
complete the square by taking half of the coefficient of the x term then squaring it.
2(x^2+x+1/4)-1/2+4 (since we added 2*1/4 to the equation we must subtract an equal amount = 1/2)
=2(x+1/2)^2+3+1/2
This is now in the form that shows the (x,y) coordinates are (-1/2,3+1/2), same as found by the first method.

ans: coordinates of vertex are (-1/2,3+1/2)
maximum value = 3+1/2
line of symmetry = x=-1/2

See below the graph of parabola

{{{ graph( 300, 200, -5, 5, -5, 10, 2x^2+2x+4) }}}