Question

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

Answer 1

`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]}