Find the degree, leading term, leading coefficient, constant term, and end behavior of the given polynomial. \( f(x)=7 x^{5}(9 x-1)(x+4)^{2} \) Degree: Leading term: Leading coefficient: End behavior: As \( x \rightarrow-\infty, f(x) \rightarrow \square \) As \( x \rightarrow \infty, f(x) \rightarrow \square \)

To analyze the polynomial \( f(x) = 7 x^{5}(9 x - 1)(x + 4)^{2} \), let's break it down: The degree of the polynomial is determined by the highest power of \( x \). In this case: - \( 7 x^5 \) contributes 5, - \( (9x - 1) \) contributes 1, - \( (x + 4)^{2} \) contributes 2 (as it is a square). Adding these together: \( 5 + 1 + 2 = 8 \). Thus, the degree is 8. The leading term is formed by multiplying the leading coefficients of each factor: - From \( 7x^5 \), we get \( 7x^5 \). - From \( (9x - 1) \), we take \( 9x \). - From \( (x + 4)^{2} \), the leading term is \( x^2 \). So, the leading term is \( 7 * 9 * x^5 * x * x^2 = 63x^8 \). The leading coefficient, the coefficient of the leading term, is 63. The constant term will be obtained when \( x = 0 \) in \( f(x) \): Plugging in, we get \( f(0) = 7 \times 0^{5} \times (9 \times 0 - 1) \times (0 + 4)^{2} = 7 \times 0 \times (-1) \times 16 = 0\). For end behavior, since the polynomial is of even degree (8) with a positive leading coefficient (63), we can conclude: As \( x \rightarrow -\infty, f(x) \rightarrow \infty \) As \( x \rightarrow \infty, f(x) \rightarrow \infty \) So, summarizing: Degree: 8 Leading term: \( 63x^8 \) Leading coefficient: 63 Constant term: 0 End behavior: As \( x \rightarrow -\infty, f(x) \rightarrow \infty \) As \( x \rightarrow \infty, f(x) \rightarrow \infty \)