Question

What is the vertex form of `y=x^2 - 19x +14`?

Answer 1

The vertex form is `(x-k)^2=4p(y-k)`
`(x-19/2)^2=y--305/4`
with vertex at `(h, k)=(19/2, (-305)/4)`

Explanation:

Start from the given equation `y=x^2-19x+14`

Divide 19 by 2 then square the result to obtain `361/4`. Add and subtract `361/4` to the right side of the equation right after `-19x`

`y=x^2-19x+14`

`y=x^2-19x+361/4-361/4+14`

the first three terms form a PERFECT SQUARE TRINOMIAL

`y=(x^2-19x+361/4)-361/4+14`

`y=(x-19/2)^2-361/4+14`

`y=(x-19/2)^2-361/4+56/4`

`y=(x-19/2)^2-305/4`

`y--305/4=(x-19/2)^2`

`(x-19/2)^2=y--305/4`

God bless....I hope the explanation is useful.