## Calculus (3rd Edition)

$y=-17$ is the minimum, when $x=-2$
Given $$y=3x^{2}+12 x -5$$ So, we have \begin{aligned} y&=3x^{2}+12 x -5 \\ &=3\left(x^{2}+4 x\right)-5\\ & =3\left(x^{2}+4 x+4-4\right)-5\\ & =3\left(x^{2}+4 x+4\right)-5-12\\ & =3\left(x+2\right)^{2}-17\\ \end{aligned} The lowest value for the squared term is $0$. So, we see that $y=-17$ is the minimum, when $x=-2$.