Answer
(a) $3$.
(b) $\pm1,\pm3,\pm5,\pm15,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{5}{2},\pm\frac{15}{2}$.
(c) $3,-5,-\frac{1}{2}$. $g(x)=(2x+1)(x+5)(x-3)$
(d) $x=-5,-\frac{1}{2}, 3$. $g(0)=-15$.
(e) crosses the x-axis at each x-intercept.
(f) $y=2x^3$
(g) See graph.
Work Step by Step
Given $g(x)=2x^3+5x^2-28x-15$, we have:
(a) As the degree $n=3$, we can find the maximum number of real zeros as $3$.
(b) We can list the potential rational zeros as $\pm1,\pm3,\pm5,\pm15,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{5}{2},\pm\frac{15}{2}$.
(c) Use synthetic division to find a real zeros $x=3$ as shown in the figure. Use the quotient and solve $2x^2+11x+5=0$ to get $(2x+1)(x+5)=0$ and $x=-5,-\frac{1}{2}$. Thus $g(x)=(2x+1)(x+5)(x-3)$
(d) To find the x-intercept(s), let $y=0$ to get $x=-5,-\frac{1}{2}, 3$. To find the y-intercept(s), let $x=0$ to get $g(0)=-15$.
(e) For $x=-5,-\frac{1}{2}, 3$, each has a multiplicity of 1 and the graph crosses the x-axis at each x-intercept.
(f) The power function that the graph resembles can be found as $y=2x^3$ for large values of $|x|$ .
(g) See graph.
