Answer
Minimum at $(2,1)$
Work Step by Step
Step 1. Given the function $y=2x^2-8x+9$, take the derivative to get $y'=4x-8$
Step 2. The critical points can be found when $g'(x)=0$ or is undefined; we have $4x-8=0$, which gives $x=2$
Step 3. An extreme value is at $x=2, y=2(2)^2-8(2)+9=1$
Step 4. As $x\to\pm\infty, y\to\infty$, we can identify that $(2,1)$ is an absolute and local minimum as shown in the figure.
