Question

How do you find the vertex of the parabola: `y=x^2+2x+2`?

Answer 1

Vertex: `(-1,1)`

Explanation:

There are two methods to solve this:

Method 1 : Converting to Vertex Form
Vertex form can be represented as `y=(x-h)^2+k`
where the point `(h,k)` is the vertex.

To do that, we should complete the square
`y=x^2+2x+2`
First, we should try to change the last number in a way
so we can factor the entire thing
`=>` we should aim for `y=x^2+2x+1`
to make it look like `y=(x+1)^2`

If you notice, the only difference between the original `y=x^2+2x+2` and the factor-able `y=x^2+2x+1` is simply changing the `2` to a `1`
[Since we can't randomly change the 2 to a 1, we can add 1 and subtract a 1 to the equation at the same time to keep it balanced.]

[ So we get... ] `y=x^2+2x+1+2-1`
[ Organizing... ] `y=(x^2+2x+1)+2-1`
[ Add like terms.. 2-1=1 ] `y=(x^2+2x+1)+1`
[ Factor! :) ] `y=(x+1)^2+1`
Now comparing it to `y=(x-h)^2+k`
We can see that the vertex would be `(-1,1)`

-----.:.-----

Method 2 : Axis of Symmetry

The axis of symmetry of a quadratic equation aka parabola is represented by `x={-b}/{2a}` when given `y=ax^2+bx+c`

Now in this case of `y=x^2+2x+2`,
we can determine that `a=1`, `b=2`, and `c=2`

plugging this into the `x=-b/{2a}`
we get `-2/{2*1}=-2/2=-1`
therefore the x point of the vertex would be `-1`

to find the y point of the vertex all we have to do is plug `x=-1` back into the `y=x^2+2x+2` equation

we would get: `y=(-1)^2+2(-1)+2`
simplify: `y=1-2+2 = 1`
therefore the y point of the vertex would be `1`

with these two pieces of information, `(x,y)`
would become `(-1,1)` which would be your vertex :)