What is the vertex form of #y= x^2+x/2-4 #?

1 Answer
Dec 21, 2015

#y=1(x-(-1/4))^2+(-4 1/16)#

Explanation:

Given:
#color(white)("XXX")y=x^2+x/2-4#

Complete the square:
#color(white)("XXX")y=x^2+1/2xcolor(green)(+(1/4)^2) -4 color(green)(-(1/4)^2)#

Re-write as a squared binomial plus a simplified constant:
#color(white)("XXX")y=(x+1/4)^2- 4 1/16#

Complete vertex form is #y=m(x-a)^2+b#
so we adjust signs to get this form (an include the default value for #m#)
#color(white)("XXX")y=1(x-(-1/4))^2+(-4 1/16)#
which has its vertex at #(-1/4,-4 1/16)#
graph{x^2+x/2-4 [-3.813, 6.054, -4.736, 0.196]}