Answer
See graph.
Work Step by Step
Step 1. For $f(x)=2x^2-4x+1$, we have $a=2\gt0, b=-4$, we can determine the graph opens up
Step 2. We can find its vertex at $x=-\frac{b}{2a}=-\frac{-4}{2(2)}=1$ and $f(1)=2-4+1=-1$ or $(1,-1)$,
Step 3. We can find the axis of symmetry $x=1$,
Step 4. We can find the y-intercept $y=1$, x-intercept(s) $x=\frac{4\pm\sqrt {16-8}}{2(2)}==\frac{2\pm\sqrt {2}}{2}$.
Step 5. See graph.
