Answer
(a) $n=3$.
(b) $\pm1,\pm3,\pm5,\pm15,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{5}{2},\pm\frac{15}{2}$.
(c) $-5,-\frac{1}{2},3$, thus $g(x)=(x-3)(x+5)(2x+1)$.
(d) $x=-5,-\frac{1}{2},3$, $f(0)=-15$.
(e) $x=-5,-\frac{1}{2},3$ crosses the x-axis.
(f) $y=2x^3$
(g) See graph.
Work Step by Step
(a) Given $g(x)=2x^3+5x^2-28x-15$, we can determine the maximum number of real zeros as $n=3$.
(b) We can list the potential rational zeros $\pm1,\pm3,\pm5,\pm15,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{5}{2},\pm\frac{15}{2}$.
(c) Use synthetic division, we can find a zero $x=3$ and quotient $2x^2+11x+5=(x+5)(2x+1)$ which gives two more zeros $x=-5,-\frac{1}{2}$, thus $g(x)=(x-3)(x+5)(2x+1)$.
(d) We can find the x-intercepts $x=-5,-\frac{1}{2},3$, y-intercepts $f(0)=-15$.
(e) At $x=-5,-\frac{1}{2},3$ the graph crosses the x-axis.
(f) The power function that the graph resembles is $y=2x^3$ for end behaviors.
(g) See graph.
