What is the vertex form of #y=(5x-5)(x+20)#?

1. Expand.
Rewrite the equation in standard form.

#y=(5x-5)(x+20)#

#y=5x^2+100x-5x-100#

#y=5x^2+95x-100#

2. Factor 5 from the first two terms.

#y=5(x^2+19x)-100#

3. Turn the bracketed terms into a perfect square trinomial.
When a perfect square trinomial is in the form #ax^2+bx+c#, the #c# value is #(b/2)^2#. So you have to divide #19# by #2# and square the value.

#y=5(x^2+19x+(19/2)^2)-100#

#y=5(x^2+19x+361/4)-100#

4. Subtract 361/4 from the bracketed terms.
You can't just add #361/4# to the equation, so you have to subtract it from the #361/4# you just added.

#y=5(x^2+19x+361/4# #color(red)(-361/4))-100#

5. Multiply -361/4 by 5.
You then need to remove the #-361/4# from the brackets, so you multiply it by your #a# value, #color(blue)5#.

#y=color(blue)5(x^2+19x+361/4)-100[color(red)((-361/4))*color(blue)((5))]#

6. Simplify.

#y=5(x^2+19x+361/4)-100-1805/4#

#y=5(x^2+19x+361/4)-2205/4#

7. Factor the perfect square trinomial.
The last step is to factor the perfect square trinomial. This will tell you the coordinates of the vertex.

#color(green)(y=5(x+19/2)^2-2205/4)#