Here is one way to approach converting the given form #y=2x^2-5x# into vertex form: #m(x-a)^2+b#
#y=2x^2-5x#
#rArr y/2=x^2-5/2x#
Remember that the expansion perfect square binomial will have the form #x^2+2ax+a^2#
If #x^2-5/2x# are the first two terms of the expansion of a perfect square binomial, then
#color(white)("XXX")a=-5/4#
and
#color(white)("XXX")a^2=(-5/4)^2#
So we will need to add #(-5/4)^2# to "complete the square" (of course if we add this to one side we will also need to add it to the other to keep the equation valid).
#y/2+(-5/4)^2=x^2-5/2x+(-5/4)^2#
#color(white)("XXXXXXXX")=(x-5/4)^2#
To get this into its proper form we need to isolate the #y# on the left side:
Subtracting #(-5/4)^2=25/16# from both sides:
#color(white)("XXX")y/2=(x-5/4)^2-(25/16)#
then multiplying both sides by #2#
#color(white)("XXX")y=2(x-5/4)^2+(-25/8)#
Here is the graph of #y=2x^2-5x# to indicate that this result is reasonable:
graph{2x^2-5x [-3.03, 5.742, -4.165, 0.217]}
