Determine the vertex of the function 2x^2-7x+5

Let the function be y(x) =
2x^2-7x+5.

To determine the vertex, we first bring this
into the standard form of a parabola like  (x-h)^2 = 4a(y-k), where  (h,k) are the
vertex of the parabola, a is focal distance from the origin along the y
axis.

y = 2x^2-7x +5 .

We
divide both sides by 2.

y/2 =
x^2-(7/2)x+5/2.

We write right as (x^2-7/4 )^2 - (7/4)^2
+5/2.

y/2 = (x-7/4)^2 -  49/16
+40/16.

y/2 = (x-7/4)^2
-9/16.

y/2 +9/16 =
(x-7/4)^2.

(y + 9/8)/2  =
(x-7/4)^2.

(x-7/4)^2 = 
(y+9/8)/2.

(x-7/4)^2 =
4*(1/8)(y+9/8).

Comparing the above with the standard
parabola (x-h)^2= 4a(y-k), we see that  the coordinates of the vetex (h, k) = (7/4 ,
-9/8).