Question

How do you convert `y=x^2+18x+95` in vertex form?

Answer 1

`y=(x+9)^2+14`

Explanation:

First find the vertex using the formula
`x=(-b)/"2a"`

`a=1`
`b=18`
`c=95`

`x=(-(18))/"2(1)"` This simplifies to `x=-18/"2"` which is `-9`.
so `x=-9`

So on now that we have `x` we can find `y`.

`y=x^2+18x+95`
`y=(-9)^2+18(-9)+95`
`y=14`

Vertex = `(-9,14)` where `h=-9` and `k=14`

We now finally enter this into vertex form which is,
`y=a(x-h)^2+k`

`x` and `y` in the "vertex form" are not associated with the values we found earlier.

`y=1(x-(-9))^2+14`
`y=(x+9)^2+14`

Answer 2

`(x+9)^2+14`

Explanation:

Firstly, you have to find the vertex point. Use the formula `x_v=-b/(2a)`. You get `(h) x_v=-9` and `(k)y_v=14`

Since there's an imaginary 1 in front of x, the a value is 1.
Now just plug everything into the equation `y=a(x-h)^2+k`