Question

for the polynomial function f(x)= - 3x^4 - 9x^3, answer the parts a through e.
a. use the leading coefficient test to determine the graphs end - behavior.
a. the graph of f(x) falls to the left and falls to the right.
b. the graph of f(x) falls to the left and rises to the right.
c. the graph of f(x) rises to the left and falls to the right.
d. the graph of f(x) rises to the left and rises to the right.
b. find the x - intercepts. state whether the graph crosses the x - axis, or touches the x - axis and turns around, at each intercept.
the x - intercept(s) is/are .
(type an integer or a decimal. use a comma to separate answers as needed. type each answer only once.)

Explanation:

Step1: Identify the leading - coefficient and degree

The polynomial function is $f(x)=-3x^{4}-9x^{3}$. The degree $n = 4$ (even) and the leading - coefficient $a=-3$ (negative).

Step2: Apply the Leading - Coefficient Test

For a polynomial $y = a_nx^n+\cdots+a_0$, when $n$ is even and $a_n<0$, as $x\to-\infty$, $y\to-\infty$ and as $x\to+\infty$, $y\to-\infty$. So the graph of $f(x)$ falls to the left and falls to the right.

Step3: Find the x - intercepts

Set $f(x)=0$, so $-3x^{4}-9x^{3}=0$. Factor out $-3x^{3}$: $-3x^{3}(x + 3)=0$.

Step4: Solve for x

Using the zero - product property, if $-3x^{3}(x + 3)=0$, then $-3x^{3}=0$ or $x + 3=0$.
From $-3x^{3}=0$, we get $x = 0$; from $x+3=0$, we get $x=-3$.
For $x = 0$, the factor is $x^{3}$, and since the exponent of the factor $x$ is 3 (odd), the graph crosses the $x$ - axis at $x = 0$. For $x=-3$, the factor is $(x + 3)$ with exponent 1 (odd), the graph crosses the $x$ - axis at $x=-3$.

Answer:

a. A. The graph of f(x) falls to the left and falls to the right.
b. - 3,0