In Problems 45-56, find the vertical, horizontal, and oblique asymptotes, if any, of each rational function. \( \begin{array}{llll}\text { 45. } R(x)=\frac{3 x}{x+4} & \text { 46. } R(x)=\frac{3 x+5}{x-6} & \text { 47. } H(x)=\frac{x^{3}-8}{x^{2}-5 x+6} & \text { 48. } G(x)=\frac{x^{3}+1}{x^{2}-5 x-14}\end{array} \)

Let's analyze each rational function to determine their vertical, horizontal, and oblique asymptotes. --- ### **Problem 45** **Function:** \( R(x) = \frac{3x}{x + 4} \) - **Vertical Asymptote:** - Set the denominator equal to zero: \( x + 4 = 0 \) ⇒ \( x = -4 \). - **Horizontal Asymptote:** - Compare the degrees of the numerator and the denominator. Both are degree 1. - The horizontal asymptote is the ratio of the leading coefficients: \( y = \frac{3}{1} = 3 \). - **Oblique Asymptote:** - None, since the degrees of the numerator and denominator are equal. **Summary:** - **Vertical Asymptote:** \( x = -4 \) - **Horizontal Asymptote:** \( y = 3 \) - **Oblique Asymptote:** None --- ### **Problem 46** **Function:** \( R(x) = \frac{3x + 5}{x - 6} \) - **Vertical Asymptote:** - Set the denominator equal to zero: \( x - 6 = 0 \) ⇒ \( x = 6 \). - **Horizontal Asymptote:** - Both numerator and denominator are degree 1. - The horizontal asymptote is the ratio of the leading coefficients: \( y = \frac{3}{1} = 3 \). - **Oblique Asymptote:** - None, since the degrees of the numerator and denominator are equal. **Summary:** - **Vertical Asymptote:** \( x = 6 \) - **Horizontal Asymptote:** \( y = 3 \) - **Oblique Asymptote:** None --- ### **Problem 47** **Function:** \( H(x) = \frac{x^3 - 8}{x^2 - 5x + 6} \) - **Vertical Asymptotes:** - Factor the denominator: \( x^2 - 5x + 6 = (x - 2)(x - 3) \). - Factor the numerator: \( x^3 - 8 = (x - 2)(x^2 + 2x + 4) \). - After cancellation, \( H(x) = \frac{x^2 + 2x + 4}{x - 3} \) with \( x \neq 2 \). - **Vertical Asymptote:** \( x = 3 \). - **Horizontal/Oblique Asymptote:** - The simplified numerator \( x^2 + 2x + 4 \) is degree 2, and the denominator \( x - 3 \) is degree 1. - Since the numerator's degree is one higher, there is an **oblique asymptote**. - Perform polynomial division: \[ \frac{x^2 + 2x + 4}{x - 3} = x + 5 + \frac{19}{x - 3} \] - **Oblique Asymptote:** \( y = x + 5 \). - **Horizontal Asymptote:** - None, since an oblique asymptote exists. **Summary:** - **Vertical Asymptote:** \( x = 3 \) - **Oblique Asymptote:** \( y = x + 5 \) - **Horizontal Asymptote:** None --- ### **Problem 48** **Function:** \( G(x) = \frac{x^3 + 1}{x^2 - 5x - 14} \) - **Vertical Asymptotes:** - Factor the denominator: \( x^2 - 5x - 14 = (x - 7)(x + 2) \). - The numerator \( x^3 + 1 = (x + 1)(x^2 - x + 1) \) shares no common factors with the denominator. - **Vertical Asymptotes:** \( x = 7 \) and \( x = -2 \). - **Horizontal/Oblique Asymptote:** - The numerator \( x^3 + 1 \) is degree 3, and the denominator \( x^2 - 5x - 14 \) is degree 2. - Since the numerator's degree is one higher, there is an **oblique asymptote**. - Perform polynomial division: \[ \frac{x^3 + 1}{x^2 - 5x - 14} = x + 5 + \frac{39x + 71}{x^2 - 5x - 14} \] - **Oblique Asymptote:** \( y = x + 5 \). - **Horizontal Asymptote:** - None, since an oblique asymptote exists. **Summary:** - **Vertical Asymptotes:** \( x = 7 \) and \( x = -2 \) - **Oblique Asymptote:** \( y = x + 5 \) - **Horizontal Asymptote:** None --- ### **Final Summary:** 1. **Problem 45:** - **Vertical Asymptote:** \( x = -4 \) - **Horizontal Asymptote:** \( y = 3 \) - **Oblique Asymptote:** None 2. **Problem 46:** - **Vertical Asymptote:** \( x = 6 \) - **Horizontal Asymptote:** \( y = 3 \) - **Oblique Asymptote:** None 3. **Problem 47:** - **Vertical Asymptote:** \( x = 3 \) - **Oblique Asymptote:** \( y = x + 5 \) - **Horizontal Asymptote:** None 4. **Problem 48:** - **Vertical Asymptotes:** \( x = 7 \) and \( x = -2 \) - **Oblique Asymptote:** \( y = x + 5 \) - **Horizontal Asymptote:** None --- Feel free to ask if you need further explanations or assistance with other problems!