(I assumed the second term was -12x and not just -12 as given)

To find the vertex form, you apply the method of:

"completing the square".

This involves adding the correct value to the quadratic expression to create a perfect square.

Recall: #(x-5)^2 = x^2 color(tomato)(-10)xcolor(tomato)(+25)" "larr color(tomato)(((-10)/2)^2 = 25)#

This relationship between #color(tomato)(b and c)# will always exist.

If the value of #c# is not the correct one, add on what you need. (Subtract it as well to keep the value of the expression the same)

#y = x^2 color(tomato)(-12)x+34" "larr ((-12)/2)^2=36 !=34#

Adding 2 will make the 36 that is needed.

#y = x^2 color(tomato)(-12)x+34 color(blue)(+2-2)" "larr# the value is the same

#y = x^2 color(tomato)(-12)x+color(tomato)(36) color(blue)(-2)#

#y = (x-6)^2-2" "larr# this is vertex form

The vertex is at #(6,-2)" "larr# note the signs

How do you get to it?

#y = color(lime)(x^2) color(tomato)(-12)x+36 color(blue)(-2)#

#y = (color(lime)(x)color(tomato)(-6))^2color(blue)(-2)#

#color(lime)(x = sqrt(x^2)) and color(tomato)((-12)/2 = -6) " check" sqrt36 = 6 #