Question

What is the vertex form of `y= - x^2 - 17x - 15`?

Answer 1

`y=-1(x+17/4)^2+57 1/4`

Explanation:

Given -

`y=-x^2-17x-15`

Find the vertex -

`x=(-b)/(2a)=(-(-17))/(2 xx(-1))=17/(-2)=(-17)/2`

`y=-((-17)/2)^2-17((-17)/2)-15`
`y=-(72 1/4)+144 1/2-15`
`y=-72 1/4+144 1/2-15`
`y=57 1/4`

Vertex is `(-17/2, 57 1/4)`

The vertex form of the quadratic equation is -

`y=a(x-h)^2+k`

Where -

`a=-1` Coefficient of `x^2`
`h=-17/4` `x` co-ordinate of the vertex
`k=57 1/4` `y` co-ordinate of the vertex

Now substitute these values in the vertex formula.

`y=-1(x-(-17/4))^2+(57 1/4)`

`y=-1(x+17/4)^2+57 1/4`

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