\begin{tabular}{l} Determine if the function below has a horizontal or a slant asymptote, and \\ explain why. \\ \( \qquad f(x)=\frac{15 x^{3}-25 x^{2}+9}{-2 x-2+3 x^{2}} \) \\ Answer Attempt 1 out of 2 \\ The function has a \\ numerator is \( \square \) asymptote because the degree of the \\ As \( x \) approaches infinity (positive or negative), the value of \( y \) approaches \\ \( \square \) the degree of the denominator. \\ \hline . \end{tabular}

Let's determine whether the function \( f(x) = \frac{15x^{3} - 25x^{2} + 9}{3x^{2} - 2x - 2} \) has a horizontal or a slant (oblique) asymptote by analyzing the degrees of the numerator and the denominator. ### Step 1: Compare Degrees of Numerator and Denominator - **Degree of the numerator:** 3 (the highest power of \( x \) in the numerator is \( x^3 \)) - **Degree of the denominator:** 2 (the highest power of \( x \) in the denominator is \( x^2 \)) ### Step 2: Determine the Type of Asymptote - **If the degree of the numerator is equal to the degree of the denominator**, there is a **horizontal asymptote** at \( y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} \). - **If the degree of the numerator is exactly one more than the degree of the denominator**, there is a **slant (oblique) asymptote**. This asymptote can be found by performing polynomial long division. - **If the degree of the numerator is more than one higher than the degree of the denominator**, there is **no horizontal or slant asymptote**. In this case, the **degree of the numerator (3)** is **one higher than the degree of the denominator (2)**. Therefore, the function has a **slant asymptote**. ### Step 3: Find the Slant Asymptote Perform polynomial long division of the numerator by the denominator: \[ \frac{15x^{3} - 25x^{2} + 9}{3x^{2} - 2x - 2} = 5x - 5 + \frac{-1}{3x^{2} - 2x - 2} \] The **slant asymptote** is the linear part of the division: \[ y = 5x - 5 \] ### Final Answer with Filled Blanks \[ \begin{tabular}{l} The function has a \textbf{slant} asymptote because the degree of the \textbf{numerator} is one higher than the degree of the denominator. \\ As \( x \) approaches infinity (positive or negative), the value of \( y \) approaches the slant asymptote \( y = 5x - 5 \), since the degree of the numerator is one higher than the degree of the denominator. \end{tabular} \]