Answer
$ y=-9$ is the minimum, when $ x=1$
Work Step by Step
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 $x=1$.
