What is the vertex form of #y=13x^2 +3x- 36 #?

1. Factor 13 from the first two terms.

#y=13x^2+3x-36#

#y=13(x^2+3/13x)-36#

2. 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#. Thus you divide #3/13# by #2# and square the value.

#y=13(x^2+3/13x+(3/13x-:2)^2)-36#

#y=13(x^2+3/13x+9/676)-36#

3. Subtract 9/676 from the perfect square trinomial.
You cannot just add #9/676# to the equation, so you must subtract it from the #9/676# you just added.

#y=13(x^2+3/13x+9/676# #color(red)(-9/676))-36#

4. Multiply -9/676 by 13.
The next step is to bring #-9/676# out of the brackets. To do this, multiply #-9/676# by the #a# value, #13#.

#y=color(blue)13(x^2+3/13x+9/676)-36[color(red)((-9/676))*color(blue)((13))]#

5. Simplify.

#y=(x^2+3/13x+9/676)-36-9/52#

#y=(x^2+3/13x+9/676)-1881/52#

6. Factor the perfect square trinomial.
The last step is to factor the perfect square trinomial. This will allow you to determine the coordinates of the vertex.

#color(green)(y=(x+3/26)^2-1881/52)#

#:.#, the vertex form is #y=(x+3/26)^2-1881/52#.