Question

answer parts a.– e. for the function shown below.
f(x) = 6x² - x³

a. use the leading coefficient test to determine the graph’s end behavior.
which statement describes the end behavior of f(x)?
○ a. the graph of f(x) rises left and falls right.
○ b. the graph of f(x) falls left and rises right.
○ c. the graph of f(x) falls left and falls right.
○ d. the graph of f(x) rises left and rises right.

Explanation:

Step1: Identify Degree and Leading Coefficient

The function is \( f(x) = 6x^2 - x^3 \), which can be rewritten as \( f(x)= -x^3 + 6x^2 \). The degree (highest power of \( x \)) is 3 (odd), and the leading coefficient (coefficient of the highest - degree term) is - 1 (negative).

Step2: Apply Leading Coefficient Test

For a polynomial function \( f(x)=a_nx^n + a_{n - 1}x^{n - 1}+\cdots+a_1x + a_0 \):

• If the degree \( n \) is odd:
• If the leading coefficient \( a_n>0 \), as \( x

ightarrow+\infty \), \( f(x)
ightarrow+\infty \) (rises to the right) and as \( x
ightarrow-\infty \), \( f(x)
ightarrow-\infty \) (falls to the left).

• If the leading coefficient \( a_n < 0 \), as \( x

ightarrow+\infty \), \( f(x)
ightarrow-\infty \) (falls to the right) and as \( x
ightarrow-\infty \), \( f(x)
ightarrow+\infty \) (rises to the left).

Since our polynomial has an odd degree (\( n = 3 \)) and a negative leading coefficient (\( a_n=- 1<0 \)), as \( x
ightarrow+\infty \) (right - hand end), \( f(x)
ightarrow-\infty \) (falls right), and as \( x
ightarrow-\infty \) (left - hand end), \( f(x)
ightarrow+\infty \) (rises left).

Answer:

A. The graph of \( f(x) \) rises left and falls right.