## Calculus (3rd Edition)

$y=-9$ is the minimum, when $x=1$
Given $$y= 2 x^{2}-4 x-7$$ So, we have \begin{aligned} y&=2 x^{2}-4 x-7 \\ &=2\left(x^{2}-2 x\right)-7\\ & =2\left(x^{2}-2 x+1-1\right)-7\\ & =2\left((x-1)(x-1)-1\right)-7\\ & =2\left((x-1)^2-1\right)-7\\ & =2(x-1)^2-9\\ \end{aligned} The lowest value for the squared term is $0$. So, we see that $y=-9$ is the minimum, when $x=1$.