Which of the following polynomials could have the same end behavior as f(x)=ax6+bx5+c?

There may be more than one correct answer. Select all correct answers.

A)nx5+bx4+c
B)−jx8+bx7+cx4
C)kx+b
D)−mx2
E)dx4−bx3−cx2+dx+e

I think it is A and D

To determine which of the polynomials could have the same end behavior as f(x) = ax^6 + bx^5 + c, we need to analyze the leading term. The leading term of a polynomial determines its end behavior.

In this case, the leading term of f(x) is ax^6. The degree of the polynomial is 6, and the coefficient of the leading term is a.

For two polynomials to have the same end behavior, they must have the same degree and the same sign on the leading term coefficient.

Now let's analyze each option:

A) nx^5 + bx^4 + c: The leading term is nx^5. Since n is a constant and not equal to a, the leading term coefficient is different, so it does not have the same end behavior as f(x).

B) -jx^8 + bx^7 + cx^4: The leading term is -jx^8. Since the degree is different (8 ≠ 6) and the leading term coefficient is negative (-j ≠ a), it does not have the same end behavior as f(x).

C) kx + b: The leading term is kx (degree = 1). Since the degree is different (1 ≠ 6), it does not have the same end behavior as f(x).

D) -mx^2: The leading term is -mx^2. Since the degree is different (2 ≠ 6) and the leading term coefficient is negative (-m ≠ a), it does not have the same end behavior as f(x).

E) dx^4 - bx^3 - cx^2 + dx + e: The leading term is dx^4. Since the degree is the same (4 = 6), we need to compare the leading term coefficients. Since the coefficient of dx^4 is not equal to a, it does not have the same end behavior as f(x).

Based on the analysis, none of the offered polynomials (A), (B), (C), (D), or (E) have the same end behavior as f(x) = ax^6 + bx^5 + c. Therefore, the correct answer is none of the above options.

no odd-degree polynomial has the same end behavior as any even-degree polynomial.

all even-degree polynomials have the same end behavior, if their leading coefficient has the same sign.