F(x)=4+5x+x2 how to identify vertex

Please type "x^2" to indicate exponentiation.  Your function is

F(x) = 4 + 5x + x^2.  If written in "standard form," it's F(x) = x^2 + 5x + 4.

This has the configuration y = ax^2 + bx + c.  In your F(x), a=1, b=5 and c=4.

The x-coordinate of the vertex is x=-b/(2a).  Using a=1 and b=5, this x-coordinate is x= -5/(2[a]), or x = -5/2.

Find the y-coordinate of the vertex by plugging x=-5/2 into F(x)=x^2+5x+4:

F(-5/2) = (-5/2)^2 + 5(-5/2) + 4 = (25/4) - (25/2) + 4.

Simplifying this, using the LCD, we get (25/4) - (50/4) + (16/4).

Then F(-5/2) = -9/4.

The vertex is at (-5/2, -9/4).

There are other ways to find the vertex.  Let me know if you want to discuss this problem further.