Answer
(a) $x=0$ (multiplicity 2), $x=-\sqrt 2,\sqrt 2$ (multiplicity 1).
(b) crosses the x-axis at $x=\pm\sqrt 2$, touches at $x=0$.
(c) $3$.
(d) $y=-2x^4$.
Work Step by Step
(a) Given $f(x)=-2x^2(x^2-2)=-2x^2(x+\sqrt 2)(x-\sqrt 2)$, we can find zero(s) $x=0$ (multiplicity 2), $x=-\sqrt 2$ (multiplicity 1), $x=\sqrt 2$ (multiplicity 1).
(b) The graph crosses the x-axis at $x=\pm\sqrt 2$, touches at $x=0$.
(c) The maximum number of turning points on the graph is $n-1=4-1=3$.
(d) The end behavior is similar to those of $y=-2x^4$.
