Answer
The function has a minimum value of $1$.
Work Step by Step
The given function is
$\Rightarrow y=x^2+6x+10$
Subtract $1$ from each side.
$\Rightarrow y-1=x^2+6x+10-1$
Simplify.
$\Rightarrow y-1=x^2+6x+9$
Write the right side as the square of a binomial.
$\Rightarrow y-1=(x+3)^2$
Write in vertex form.
$\Rightarrow y=(x+3)^2+1$
The vertex is $(-3,1)$. Because $a$ is positive $(a=1)$, the parabola opens up and the $y-$coordinate of the vertex is the minimum value.
Hence, the function has a minimum value of $1$.
