Intermediate Algebra (6th Edition)

$(-1, 5)$
The vertex-form of a quadratic function is $y=a(x-h)^{2}+k$ where the vertex is the opposite of $h$ and $k$, or $(h, k)$. Given the function from the problem, $g(x)=(x+1)^{2}+5$, written in vertex form this function looks like $f(x)=1(x-(-1))^{2}+5$. Therefore the vertex is at $(-1, 5)$.