Question

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

Answer 1

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

Explanation:

Given _

`y=x^2+2x+15`

The vertex form of the equation is -

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

If we know the values of `a,h and k` we can change the given equation into a vertex form.

Find the vertex `(h, k)`

`a` is the coefficient of `x^2`
`h` is the x-co-ordinate of the vertex
`k` is the y-co-ordinate of the vertex

`a=1`
`h= (-b)/(2a)=(-2)/(2 xx 1)=-1`

`k=(-1)^2+2(-1)+15=1-2+15=14`

Now substitute the values of `a,h and k` in the vertex form of the equation.

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

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