Answer
See graph and explanations.
Work Step by Step
Step 1. Given $f(x)=2x^2(x-2)^2$, test symmetry $f(-x)=2(-x)^2(-x-2)^2=2x^2(x+2)^2$, no symmetry.
Step 2. The leading term is $2x^4$ and we can find its end behavior as rise to the left and rise to the right.
Step 3. The zeros are $x=0$ (multiplicity 2, graph touches the x-axis and turns around) and $x=2$ (multiplicity 2, graph touches the x-axis and turns around)
Step 4. The y-intercept can be found as $f(0)=0$
Step 5. Use test points as necessary, graph the function as shown in the figure.
