Find the vertex for y= -x^2+1/2x+2

I could not figure out the vertex! Please help me!

y= -x^2 + 1/2x + 2

There are plenty of ways to get its maximum/minimum. One way is to graph and locate on the graph the point where max/min occurs.
Another is to solve for its vertex. We can transform the equation into the form
y = a(x-h)^2 + k
where (h,k) is the vertex.
Therefore,
y = -x^2 + 1/2x + 2
y = -(x^2 - 1/2x) + 2
Completing the square:
y = -(x^2 - 1/2x + 1/16) + 2 + 1/16
y = -(x - 1/4)^2 + 33/16
The vertex is therefore at (1/4 , 33/16).

Another way is to use derivatives. We get the derivative of the function with respect to x:
y = -x^2 + 1/2x + 2
y' = -2x + 1/2
Then we equate it to zero (because at max/min, the slope of the tangent line is zero):
0 = -2x + 1/2
2x = 1/2
x = 1/4
Substituting this back on the original function,
y = -x^2 + 1/2x + 2
y = -(1/4)^2 + 1/2(1/4) + 2
y = -1/16 + 1/8 + 2
y = 33/16
Thus, vertex is at (1/4 , 33/16).
By the way, this vertex is a maximum, because the numerical coefficient of x^2 in the function is negative.

hope this helps~ `u`