Question

y=x^2+2x+1 find the vertex

Answer 1

I'm going to put this into vertex form, which is y=a(x-h)^2+k

To do this, you use completing the square. Start by subtracting 1 from each side

-1+y=x^2+2x

Then, divide b (2) by 2 and square it. Add that number to each side. (normally when you complete the square, you take out the gcf of the right side of the equation before you do this. Then, you would multiply (b/2)^2 by the gcf before adding it to the left side. Since there is no gcf, we do not have to worry about this)

2/2=1
1^2=1
-1+1+y=x^2+2x+1

Next, simplify the left side and factor the right side.

y=(x+1)^2

This is vertex form. a=1, h=-1, and k=0

The vertex is (h,k) so it is (-1,0)

hope this helps!

Answer 2

To find the vertex of the given parabola formula, we use completing the square. Thus, 
y = x2 + 2x + 1The right side is a perfect squarey = (x+1)^2 The standard form of the parabola is y - k = 4a (x-h ) ^2 Hence the vertex is at (-1,0)