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

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)#

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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 :)