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Graph all asymptotes of the rational function. \[ f(x)=\frac{-x^{2}-2 x+3}{x-2} \]

Ask by Cross Adkins. in the United States
Mar 10,2025

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The function \( f(x) = \frac{-x^{2}-2x+3}{x-2} \) has a vertical asymptote at \( x = 2 \) and no horizontal or oblique asymptotes.

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To graph the asymptotes of the rational function \( f(x)=\frac{-x^{2}-2 x+3}{x-2} \), we need to identify both vertical and horizontal (or oblique) asymptotes. **Vertical Asymptote:** A vertical asymptote occurs where the denominator is zero and the numerator is not zero at that point. Setting the denominator equal to zero: \[ x - 2 = 0 \implies x = 2 \] So, there is a vertical asymptote at \( x = 2 \). **Horizontal/Oblique Asymptote:** To find the horizontal or oblique asymptote, we need to consider the degree of the numerator and the denominator. Here, the degree of the numerator (\(-x^2 - 2x + 3\)) is 2, and the degree of the denominator (\(x - 2\)) is 1. Since the degree of the numerator is greater than that of the denominator by 1, we will have an oblique asymptote. To find the oblique asymptote, we perform polynomial long division of the numerator by the denominator: 1. Divide \(-x^2\) by \(x\), which gives \(-x\). 2. Multiply \(-x\) by \(x - 2\), resulting in \(-x^2 + 2x\). 3. Subtract this from the original numerator: \[ (-x^2 - 2x + 3) - (-x^2 + 2x) = -4x + 3 \] 4. Divide \(-4x\) by \(x\), giving \(-4\). 5. Multiply \(-4\) by \(x - 2\), resulting in \(-4x + 8\). 6. Subtract again: \[ (-4x + 3) - (-4x + 8) = -5 \] Thus, we have: \[ f(x) = -x - 4 + \frac{-5}{x-2} \] As \(x\) approaches infinity, the term \(\frac{-5}{x-2}\) approaches 0. Therefore, the oblique asymptote is: \[ y = -x - 4 \] **Summary of Asymptotes:** - Vertical asymptote at \(x = 2\) - Oblique asymptote at \(y = -x - 4\) You can now sketch the function, marking the vertical asymptote and the oblique asymptote to illustrate their behavior.

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