Question

use a graphing utility to graph the function. use a -5,5 by -24,24 viewing rectangle. then find the intervals on which the function is increasing, decreasing, or constant. f(x)=x^3 - 3x^2 - 9x + 5. choose the correct graph below.

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=x^{3}-3x^{2}-9x + 5$ using the power - rule $(x^n)^\prime=nx^{n - 1}$ is $f^\prime(x)=3x^{2}-6x - 9$.

Step2: Set the derivative equal to zero

$3x^{2}-6x - 9 = 0$. Divide through by 3: $x^{2}-2x - 3=0$. Factor the quadratic equation: $(x - 3)(x+1)=0$. So the critical points are $x=-1$ and $x = 3$.

Step3: Analyze the sign of the derivative

Choose test points in the intervals $(-\infty,-1)$, $(-1,3)$ and $(3,\infty)$. Let's choose $x=-2$, $x = 0$ and $x = 4$.
For $x=-2$, $f^\prime(-2)=3\times(-2)^{2}-6\times(-2)-9=12 + 12-9 = 15>0$.
For $x = 0$, $f^\prime(0)=3\times0^{2}-6\times0 - 9=-9<0$.
For $x = 4$, $f^\prime(4)=3\times4^{2}-6\times4 - 9=48-24 - 9 = 15>0$.
The function is increasing on $(-\infty,-1)\cup(3,\infty)$ and decreasing on $(-1,3)$.
To graph the function, we can also find the $y$-intercept by setting $x = 0$, $y=f(0)=5$.
We can then use a graphing utility to graph $y=x^{3}-3x^{2}-9x + 5$ in the viewing rectangle $[-5,5]$ by $[-24,24]$.
The general shape of a cubic function $y = ax^{3}+bx^{2}+cx + d$ with $a = 1>0$ starts from the bottom - left and goes to the top - right.

Answer:

Without seeing the actual graphs A, B, C, D, we cannot choose the specific graph. But the function $f(x)=x^{3}-3x^{2}-9x + 5$ has a $y$-intercept at $(0,5)$, is increasing on $(-\infty,-1)\cup(3,\infty)$ and decreasing on $(-1,3)$. If you provide the characteristics of each graph (such as $y$-intercept, intervals of increase and decrease), we can make a selection.