Answer
(a) $x=0$ (multiplicity 2), $x=-\sqrt 2$ (multiplicity 1) and $x=\sqrt 2$ (multiplicity 1).
(b) $x=\pm\sqrt 2$ crosses the x-axis, $x=0$ touches the x-axis.
(c) $3$.
(d) $y=-2x^4$.
Work Step by Step
(a) For $f(x)=-2x^2(x^2-2)$, we can list real zero as $x=0$ (multiplicity 2), $x=-\sqrt 2$ (multiplicity 1) and $x=\sqrt 2$ (multiplicity 1).
(b) At $x=\pm\sqrt 2$ the graph crosses the x-axis, while at $x=0$ the graph touches the x-axis.
(c) The maximum number of turning points on the graph is given by $n-1=4-1=3$.
(d) As $n=4, a_4\lt0$, the end behaviors are fall to the right and fall to the left, similar to $y=-2x^4$.
