Rational Functions 0B7143

1. **Problem statement:** We have two graphs of rational functions and four options for each. We need to identify which function corresponds to each graph based on vertical and horizontal asymptotes and behavior near these asymptotes. 2. **Recall the general form and asymptotes of rational functions:** For a function $y = \frac{ax + b}{cx + d}$: - Vertical asymptote at $x$ where denominator $cx + d = 0 \Rightarrow x = -\frac{d}{c}$. - Horizontal asymptote at $y = \frac{a}{c}$ if degrees of numerator and denominator are equal. --- ### Câu 3 3. **Given graph properties:** - Vertical asymptote: $x = -1$ - Horizontal asymptote: $y = 2$ - Left branch above $y=2$ and goes to $+\infty$ near $x=-1$ - Right branch below $y=2$ and goes to $-\infty$ near $x=-1$ 4. **Check each option:** - A. $y = \frac{x + 4}{x - 1}$ - Vertical asymptote at $x=1$ (denominator zero) - Horizontal asymptote $y=\frac{1}{1}=1$ - Does not match vertical or horizontal asymptote. - B. $y = \frac{x - 2}{x - 1}$ - Vertical asymptote at $x=1$ - Horizontal asymptote $y=1$ - Does not match. - C. $y = \frac{2x + 1}{x + 1}$ - Vertical asymptote at $x=-1$ - Horizontal asymptote $y=\frac{2}{1}=2$ - Matches both asymptotes. - D. $y = \frac{x + 2}{x - 1}$ - Vertical asymptote at $x=1$ - Horizontal asymptote $y=1$ - Does not match. 5. **Check behavior near vertical asymptote for C:** - As $x \to -1^-$, denominator $\to 0^-$, numerator $2(-1)+1 = -2+1 = -1$ (negative) - So $y \to \frac{-1}{0^-} = +\infty$ (since negative over negative small number is positive large) - As $x \to -1^+$, denominator $\to 0^+$, numerator still $-1$ - So $y \to \frac{-1}{0^+} = -\infty$ - This matches the graph description. **Answer for Câu 3:** Option C. --- ### Câu 4 6. **Given graph properties:** - Vertical asymptote: $x=1$ - Horizontal asymptote: $y=1$ - Left branch above $y=1$ and goes to $+\infty$ as $x \to 1^-$ - Right branch goes to $-\infty$ just right of $x=1$ - Crosses x-axis at $x=2$ - Approaches $y=1$ from below on right 7. **Check each option:** - A. $y = \frac{x + 4}{x - 1}$ - Vertical asymptote at $x=1$ - Horizontal asymptote $y=1$ - Check zero: numerator zero at $x=-4$ (not 2) - B. $y = \frac{x + 2}{x - 1}$ - Vertical asymptote at $x=1$ - Horizontal asymptote $y=1$ - Zero at $x=-2$ (not 2) - C. $y = \frac{2x + 1}{x + 1}$ - Vertical asymptote at $x=-1$ (not 1) - Does not match. - D. $y = \frac{x - 2}{x - 1}$ - Vertical asymptote at $x=1$ - Horizontal asymptote $y=1$ - Zero at $x=2$ (matches) 8. **Check behavior near vertical asymptote for D:** - As $x \to 1^-$, denominator $\to 0^-$, numerator $1 - 2 = -1$ (negative) - $y \to \frac{-1}{0^-} = +\infty$ - As $x \to 1^+$, denominator $\to 0^+$, numerator $-1$ - $y \to \frac{-1}{0^+} = -\infty$ - Matches graph behavior. **Answer for Câu 4:** Option D. --- **Final answers:** - Câu 3: $y = \frac{2x + 1}{x + 1}$ (Option C) - Câu 4: $y = \frac{x - 2}{x - 1}$ (Option D)