Question

for the polynomial function f(x) = (x - 4)²(x + 3)(x - 1), answer parts a through e. a. use the leading coefficient test to determine the graphs end behavior. a. the graph of f(x) rises to the left and rises to the right. b. the graph of f(x) falls to the left and rises to the right. c. the graph of f(x) falls to the left and falls to the right. d. the graph of f(x) rises to the left and falls to the right.

Explanation:

Step1: Expand the polynomial (or find degree and leading coefficient)

First, we determine the degree of the polynomial. When we expand \( f(x) = (x - 4)^2(x + 3)(x - 1) \), the leading term will come from multiplying the leading terms of each factor. The leading term of \( (x - 4)^2 \) is \( x^2 \), of \( (x + 3) \) is \( x \), and of \( (x - 1) \) is \( x \). Multiplying these together: \( x^2 \cdot x \cdot x = x^{2 + 1 + 1}=x^4 \). So the degree of the polynomial is 4 (even), and the leading coefficient is 1 (positive, since the coefficient of \( x^4 \) is 1).

Step2: Apply Leading Coefficient Test

The Leading Coefficient Test states that for a polynomial \( f(x) = a_nx^n + \dots + a_1x + a_0 \):

• If the degree \( n \) is even:
• If the leading coefficient \( a_n \) is positive, the graph rises to the left and rises to the right.
• If the leading coefficient \( a_n \) is negative, the graph falls to the left and falls to the right.
• If the degree \( n \) is odd:
• If the leading coefficient \( a_n \) is positive, the graph falls to the left and rises to the right.
• If the leading coefficient \( a_n \) is negative, the graph rises to the left and falls to the right.

Here, degree \( n = 4 \) (even) and leading coefficient \( a_n = 1 \) (positive). So the graph rises to the left and rises to the right.

Answer:

A. The graph of f(x) rises to the left and rises to the right.