Answer
See graph.
Work Step by Step
Step 1. Given $f(x)=2x^2-4x+1$, we have $a=2\gt0$, thus its graph opens up
Step 2. Form a square to get $f(x)=2(x^2-2x+1)-2+1=2(x-1)^2-1$, we can identify its vertex $(1,-1)$ and axis of symmetry $x=1$.
Step 3. We can find y-intercept $f(0)=1$, and x-intercept(s) by letting $y=0$ to get $x_1=1-\frac{\sqrt 2}{2}$ and $x_2=1+\frac{\sqrt 2}{2}$
Step 4. See graph.