You can very easily go wrong on this one. There is a small detail that can easily be over looked.

Let #k# be a constant yet to be determined

Given:#" "y=1/5x^2-3/7x-16#.......(1)

#color(blue)("Build the vertex form equation")#

Write as:#" "y=1/5(x^2-color(green)(15/7)x)-16#..........(2)

#color(brown)("Note that "15/7xx1/5 = 3/7)#

Consider the #15/7 "from "15/7x#

Apply# 1/2xx15/7 = color(red)(15/14)#

At this point the right hand side will not be equal to y. This will be corrected later

In (2) substitute #color(red)(15/14)" for "color(green)(15/7)#

#1/5(x^2-color(red)(15/14)x)-16" "....................(2_a)#

Remove the #x# from #15/14x#

#1/5(x^(color(magenta)(2))-15/14)-16#

Take the power (index) of #color(magenta)(2)# outside the bracket

#1/5(x-15/14)^(color(magenta)(2))-16" "color(brown)("Note that an error comes from the "15/14#

#color(brown)("This is still not yet equal to y")#

Add the constant value of #color(red)(k)#

#1/5(x-15/14)^(color(magenta)(2))-16+color(red)(k)#

#color(green)("Now it is equal to "y)#

#y=1/5(x-15/14)^2-16+color(red)(k)#.........(3)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("To determine the value of "k)#

If we were to expand the bracket and multiply by the #1/5# we would have the extra value of #1/5xx(-15/14)^2#. The constant #k# is to counter this by removing it.

#color(brown)("Let me show you what I mean. Compare equation (1) to (3)")#

#1/5x^2-3/7x-16" " =" "y" "=" "1/5(x-15/14)^2-16+k#

#1/5x^2-3/7x-16" " =" "1/5(x^2-15/7x+(15/14)^2)-16+k#

#1/5x^2-3/7x-16" " =1/5x^2-3/7x+[1/5xx(15/14)^2]-16+k#

#1/5x^2-3/7x-16" " =1/5x^2-3/7x+[45/196]-16+k#

#cancel(1/5x^2)-cancel(3/7x)-cancel(16)" " =cancel(1/5x^2)-cancel(3/7x)+[45/196]-cancel(16)+k#

#=>0=45/196+k#

#=>color(red)(k=-45/196)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So equation (3) becomes:

#y=1/5(x-15/14)^2-16color(red)(-45/196)#.........(3)

#y=1/5(x-15/14)^2-3181/196#

#color(blue)("Thus vertex form" -> y=1/5(x-15/14)^2-3181/196)#