What is the the vertex of #y = (x -4)^2+9x-20 #?

Remember quadratic form is #y= ax^2 + bx+c " " " " " " " (1)#

The vertex form of the quadratic equation is #y= a(x-h)^2 + k " " " " " " " " " " " "(2)#

Notice the equation is #y= (x-4)^2 + 9x -20# is not in the vertex form.

We begin by expand the equation like so

#y= (x-4)^2 + 9x -20#

#y = (x-4)(x-4) + 9x - 20#
#y= (x^2 -4x-4x+16)+9x-20#
#y= x^2 -8x + 16 + 9x -20#
# y= x^2 +x -4 " " " " " " " " " " " " " " (3) #

After we simplify function, we have #y = x^2 +x-4# , we can write it in the vertex form by the process of completing the square.

#y = (x^2 + x +color(red)square) -4 -color(blue)(square) " " " " (4)#

Note: the goal of completing the square if to create a perfect trinomial.
The number in the square is #(b/2)^2#

In this case, the middle term is #1# , #(1/2)^2 = 1/4 " " " " " (5)#

#y = (x^2 + x +color(red)(1/4)) -4 -color(blue)(1/4) " " " " (6)#

#y = (x^2 + x +color(red)(1/4)) -16/4 -color(blue)(1/4)#
#y = (x +1/2)^2 -17/4 " " " " " "(7)#

There is alternative method to fid the vertex using
#x_(vertex) = -b/(2a)#

#y_(vertex) = f(-b/(2a))#

I hope this help.