Answer
$y=(x+1)^2+4$
Work Step by Step
Recall:
(1) The vertex form of $y=ax^2+bx+c$ is $y=a(x-h)^2+k$ where $(h, k)$ is the vertex.
(2) The vertex of the graph of $y=ax^2+bx+c$ is at $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$.
The given function has $a=1$ and $b=2$.
Thus, the $x$-coordinate of the vertex is:
\begin{align*}
x&=-\frac{b}{2a}\\\\
x&=-\frac{2}{2(1)}\\\\
x&=-\frac{2}{2}\\\\
x&=-1
\end{align*}
Find the $y$-coordinate of the vertex by substituting $-1$ to $x$ to obtain:
\begin{align*}
y&=x^2+2x+5\\
y&=(-1)^2+2(-1)+5\\
y&=1-2+5\\
y&=4
\end{align*}
The vertex is at $(-1, 4)$ so $h=-1$ and $k=4$.
Substitute the values of $h$ and $k$ into the vertex form in the recall part (1) above to obtain:
$$y=(x+1)^2+4$$
