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

Aug 10, 2015

Vertex: $\left(- 1 , 1\right)$

#### Explanation:

There are two methods to solve this:

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

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

If you notice, the only difference between the original $y = {x}^{2} + 2 x + 2$ and the factor-able $y = {x}^{2} + 2 x + 1$ 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... ] $y = {x}^{2} + 2 x + 1 + 2 - 1$
[ Organizing... ] $y = \left({x}^{2} + 2 x + 1\right) + 2 - 1$
[ Add like terms.. 2-1=1 ] $y = \left({x}^{2} + 2 x + 1\right) + 1$
[ Factor! :) ] $y = {\left(x + 1\right)}^{2} + 1$
Now comparing it to $y = {\left(x - h\right)}^{2} + k$
We can see that the vertex would be $\left(- 1 , 1\right)$

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

Method 2 : Axis of Symmetry

The axis of symmetry of a quadratic equation aka parabola is represented by $x = \frac{- b}{2 a}$ when given $y = a {x}^{2} + b x + c$

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

plugging this into the $x = - \frac{b}{2 a}$
we get $- \frac{2}{2 \cdot 1} = - \frac{2}{2} = - 1$
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 $y = {x}^{2} + 2 x + 2$ equation

we would get: $y = {\left(- 1\right)}^{2} + 2 \left(- 1\right) + 2$
simplify: $y = 1 - 2 + 2 = 1$
therefore the y point of the vertex would be $1$

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