Answer
$(-3,-4)$
Work Step by Step
The standard form of the given quadratic function, $f(x)=
x^2+6x+5
,$ is
\begin{array}{l}\require{cancel}
f(x)=\left( x^2+6x \right)+5
\\\\
f(x)=\left( x^2+6x+\left( \dfrac{6}{2} \right)^2 \right)+5-\left( \dfrac{6}{2} \right)^2
\\\\
f(x)=\left( x^2+6x+9 \right)+5-9
\\\\
f(x)=\left( x+3 \right)^2-4
.\end{array}
Since the vertex of $f(x)=a(x-h)^2+k$ is at $(h,k)$, then the vertex of the equation above is
\begin{array}{l}\require{cancel}
(-3,-4)
.\end{array}
