Question

How do you write `y=2x^2-5x` into vertex form?

Answer 1

`y=2(x-5/4)^2+(-25/8)`

Explanation:

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]}