Question

consider the leading term of each polynomial function. what is the end behavior of the graph?

1. $-3x^3 - x$

a. the leading term is $-3x^3$. since $n$ is odd and $a$ is negative, the end behavior is down and down.
b. the leading term is $-3x^3$. since $n$ is odd and $a$ is negative, the end behavior is up and up.
c. the leading term is $-3x^3$. since $n$ is odd and $a$ is negative, the end behavior is up and down.
d. the leading term is $-3x^3$. since $n$ is odd and $a$ is negative, the end behavior is down and up.

Explanation:

Step1: Recall End Behavior Rules

For a polynomial \( f(x) = a_nx^n + \dots + a_1x + a_0 \), end behavior is determined by leading term \( a_nx^n \).
- If \( n \) is odd:
- If \( a_n > 0 \): as \( x\to\infty \), \( f(x)\to\infty \) (up); as \( x\to-\infty \), \( f(x)\to-\infty \) (down).
- If \( a_n < 0 \): as \( x\to\infty \), \( f(x)\to-\infty \) (down); as \( x\to-\infty \), \( f(x)\to\infty \) (up)? Wait, no—wait, odd degree: sign of \( a_n \) and direction. Wait, no: let's correct. For odd \( n \):
- Positive \( a_n \): \( x\to\infty \), \( f(x)\to\infty \); \( x\to-\infty \), \( f(x)\to-\infty \) (up and down).
- Negative \( a_n \): \( x\to\infty \), \( f(x)\to-\infty \); \( x\to-\infty \), \( f(x)\to\infty \)? Wait, no—wait, no, let's take \( y = -x^3 \). As \( x\to\infty \), \( -x^3\to-\infty \) (down); as \( x\to-\infty \), \( -(-\infty)^3 = -(-\infty) = \infty \)? Wait, no: \( x\to-\infty \), \( x^3\to-\infty \), so \( -x^3\to\infty \). Wait, so for \( y = -x^3 \) (odd, \( a=-1 \)): as \( x\to\infty \), down; as \( x\to-\infty \), up? But the option a says "down and down". Wait, maybe I messed up. Wait, the leading term here is \( -3x^3 \). Let's analyze:

Degree \( n = 3 \) (odd), leading coefficient \( a = -3 \) (negative).

For odd \( n \):
- When \( a > 0 \): end behavior is up (as \( x\to\infty \)) and down (as \( x\to-\infty \))? Wait, no—wait, \( y = x^3 \): as \( x\to\infty \), \( x^3\to\infty \) (up); as \( x\to-\infty \), \( x^3\to-\infty \) (down). So up and down.

For \( y = -x^3 \): as \( x\to\infty \), \( -x^3\to-\infty \) (down); as \( x\to-\infty \), \( -(-\infty)^3 = \infty \) (up)? Wait, no—wait, \( x\to-\infty \), \( x^3 = (-\infty)^3 = -\infty \), so \( -x^3 = -(-\infty) = \infty \). So \( y = -x^3 \): as \( x\to\infty \), down; as \( x\to-\infty \), up. But the options: option a says "down and down". Wait, maybe the question's option a is wrong? Wait, no—wait, maybe I made a mistake. Wait, let's check the options again.

Wait, the problem's option a: "The leading term is \( -3x^3 \). Since \( n \) is odd and \( a \) is negative, the end behavior is down and down." Wait, that can't be. Wait, maybe the original problem's option a is correct? Wait, no—wait, let's re-express. Wait, maybe I confused the direction. Wait, for \( y = -x^3 \), let's plug in values:

- \( x = 1000 \) (large positive): \( y = - (1000)^3 = -10^9 \) (down, towards \( -\infty \)).
- \( x = -1000 \) (large negative): \( y = - (-1000)^3 = - (-10^9) = 10^9 \) (up, towards \( \infty \))? Wait, that's up. But option a says "down and down". Wait, this is a contradiction. Wait, maybe the question has a typo, or I'm misunderstanding. Wait, no—wait, maybe the degree is 3, but the leading term is \( -3x^3 \). Wait, maybe the rule is: for odd \( n \), if \( a < 0 \), then as \( x\to\infty \), \( f(x)\to-\infty \) (down); as \( x\to-\infty \), \( f(x)\to\infty \) (up)? But option a says "down and down". Wait, maybe the original problem's option a is correct, and I'm wrong. Wait, let's check the leading term: \( -3x^3 \). Let's take \( x\to\infty \): \( -3x^3 \to -\infty \) (down). \( x\to-\infty \): \( -3(-\infty)^3 = -3(-\infty) = \infty \)? Wait, no: \( (-∞)^3 = -∞ \), so \( -3*(-∞) = ∞ \). So that's up. But option a says "down and down". Wait, maybe the question's option a is incorrect, but the options given: let's check the options again.

Wait, the options:

a. leading term \( -3x^3 \), \( n \) odd, \( a \) negative: end behavior down and down.

b. up and up.

c. up and down.

d. down and up.

Wait, maybe I made a mistake in the rule. Wait, let's recall the correct end behavior for polynomials:

- **Even degree**:
- Positive leading coefficient: both ends up (up and up).
- Negative leading coefficient: both ends down (down and down).

- **Odd degree**:
- Positive leading coefficient: right end up, left end down (up and down).
- Negative leading coefficient: right end down, left end up (down and up).

Ah! There we go. So for odd degree, negative leading coefficient: as \( x\to\infty \) (right end), \( f(x)\to-\infty \) (down); as \( x\to-\infty \) (left end), \( f(x)\to\infty \) (up). So that's "down and up" (option d). Wait, but the option a says "down and down". Wait, maybe the original problem has a mistake, but let's check the options again. Wait, the problem's option a: "down and down"—but according to the rule, odd degree, negative \( a \): down (right) and up (left). So option d: "down and up". But wait, the user's question: let's re-examine. Wait, the problem is question 41, with options a-d. Let's re-express the leading term: \( -3x^3 \), degree 3 (odd), leading coefficient -3 (negative). So end behavior: as \( x\to\infty \), \( f(x)\to-\infty \) (down); as \( x\to-\infty \), \( f(x)\to\infty \) (up). So that's "down and up" (option d). But wait, the option a says "down and down"—maybe the original problem's option a is wrong, but let's check the options again. Wait, maybe I misread the leading term. Wait, the leading term is \( -3x^3 \), degree 3 (odd), coefficient -3 (negative). So according to the rule, odd degree, negative \( a \): right end down, left end up (down and up). So option d. But wait, the option a says "down and down"—maybe the question has a typo, but let's check the options again. Wait, the user's question: the options are:

a. down and down

b. up and up

c. up and down

d. down and up

So according to the correct rule, odd degree, negative \( a \): down (right) and up (left) → option d. But wait, the option a says "down and down"—maybe the original problem's option a is incorrect, but let's check the leading term. Wait, the leading term is \( -3x^3 \), degree 3 (odd), \( a = -3 < 0 \). So end behavior: right end down, left end up → "down and up" (option d). But wait, the user's question: maybe I made a mistake. Wait, no—let's confirm with an example: \( y = -x^3 \). As \( x\to\infty \), \( y\to-\infty \) (down); as \( x\to-\infty \), \( y\to\infty \) (up). So that's "down and up" (option d). So the correct answer should be d. But wait, the option a says "down and down"—maybe the original problem has a mistake, but according to the rules, it's d. Wait, but the user's question: let's check the options again. Wait, the problem's option a: "the end behavior is down and down"—but that's for even degree, negative \( a \). So maybe the question's option a is mislabeled. But according to the rules, odd degree, negative \( a \): down (right) and up (left) → option d.

Step2: Analyze Each Option

- **Option a**: Claims "down and down". But for odd degree, negative \( a \), left end should be up. Eliminate a.
- **Option b**: "up and up"—for even degree, positive \( a \). Eliminate b.
- **Option c**: "up and down"—for odd degree, positive \( a \). Eliminate c.
- **Option d**: "down and up"—matches odd degree, negative \( a \) (right end down, left end up).

Answer:

d. The leading term is \(-3x^3\). Since \( n \) is odd and \( a \) is negative, the end behavior is down and up.