Question

What is the vertex form of `y=−x^2 − 7x + 1`?

Answer 1

The Vertex Form `(x- -7/2)^2=-(y-53/4)` with vertex at `(-7/2, 53/4)`

Explanation:

We start from the given and do the "Completing the Square Method"

`y=-x^2-7x+1`

factor out the `-1` first

`y=-1*(x^2+7x)+1`

Compute the number to be added and subtracted using the numerical coefficient of x which is the 7. Divide the 7 by 2 and square the result,...that is `(7/2)^2=49/4`

`y=-1*(x^2+7x)+1`

`y=-1*(x^2+7x+49/4-49/4)+1`

the first three terms inside the parenthesis forms a PST-perfect square trinomial.

`y=-1*(x^2+7x+49/4-49/4)+1`

`y=-1*((x^2+7x+49/4)-49/4)+1`

`y=-1*((x+7/2)^2-49/4)+1`

simplify by multiplying the -1 back and removing the grouping symbol

`y=-1(x+7/2)^2+49/4+1`

`y=-1(x+7/2)^2+53/4`

`y-53/4=-1(x+7/2)^2`

Let us form the Vertex Form
`(x-h)^2=+-4p(y-k)`

`(x- -7/2)^2=-(y-53/4)`

Kindly see the graph
graph{(x- -7/2)^2=-(y-53/4)[-30,30,-15,15]}

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