explain how to sketch a graph of the function f(x) = x^3 + 2x^2 - 8x. be sure to include end - behavior…

Explanation:

Step1: Find the zeros

Factor the function $f(x)=x^{3}+2x^{2}-8x=x(x^{2}+2x - 8)=x(x + 4)(x-2)$. Set $f(x)=0$, we get $x = 0,x=-4,x = 2$.

Step2: Determine end - behavior

Since the leading term is $x^{3}$ (odd - degree and positive leading coefficient), as $x\to-\infty$, $f(x)\to-\infty$ and as $x\to+\infty$, $f(x)\to+\infty$.

Step3: Find intervals of sign

Use test points in the intervals $(-\infty,-4),(-4,0),(0,2),(2,\infty)$. For $x=-5$ in $(-\infty,-4)$, $f(-5)=(-5)(-5 + 4)(-5-2)=-35<0$. For $x=-1$ in $(-4,0)$, $f(-1)=(-1)(-1 + 4)(-1-2)=9>0$. For $x = 1$ in $(0,2)$, $f(1)=(1)(1 + 4)(1-2)=-5<0$. For $x = 3$ in $(2,\infty)$, $f(3)=(3)(3 + 4)(3-2)=21>0$. So $f(x)$ is negative on $(-\infty,-4)\cup(0,2)$ and positive on $(-4,0)\cup(2,\infty)$.

Step4: Sketch the graph

Plot the zeros $x=-4,0,2$. Draw the curve going down to the left (as $x\to-\infty,f(x)\to-\infty$) and up to the right (as $x\to+\infty,f(x)\to+\infty$), crossing the x - axis at the zeros and being positive or negative in the appropriate intervals.

Answer:

First, factor to find zeros at $x=-4,0,2$. Determine end - behavior as $x\to-\infty,f(x)\to-\infty$ and $x\to+\infty,f(x)\to+\infty$. Find intervals of sign: negative on $(-\infty,-4)\cup(0,2)$ and positive on $(-4,0)\cup(2,\infty)$. Then sketch the graph with these features.