## Calculus (3rd Edition)

$y=-7$ is the minimum, when $x=-3$
Given $$y= x^{2}+6 x+2$$ So, we have \begin{aligned} y&= x^{2}+6 x+2 \\ &= x^{2}+6 x+2+7-7 \\ &= x^{2}+6 x+9-7 \\ &=( x^{2}+6 x+9)-7 \\ &=( x+3)(x+3)-7 \\ &=( x+3)^2-7 \\ \end{aligned} The lowest value for the squared term is $0$. So, we see that $y=-7$ is the minimum, when $x=-3$.