1. **Problem statement:** Given the function $f(x) = \frac{2x^2 - 1}{x^2 - 5x + 6}$, we need to analyze its domain, limits, asymptotes, derivatives, and graph. 2. **Condition for $f(x)$ to be defined:** The denominator cannot be zero. So, solve $x^2 - 5x + 6 = 0$. 3. **Find domain:** Factor denominator: $x^2 - 5x + 6 = (x - 2)(x - 3)$. So, $x \neq 2$ and $x \neq 3$. Domain is $\mathbb{R} \setminus \{2,3\}$. 4. **Evaluate limits at boundaries:** - As $x \to 2$, numerator $= 2(2)^2 - 1 = 8 - 1 = 7$. - Denominator $\to 0$, check sign: - For $x \to 2^-$, denominator $= (2^- - 2)(2^- - 3) \approx (\text{small negative})(\text{negative}) = \text{positive}$. - For $x \to 2^+$, denominator $= (2^+ - 2)(2^+ - 3) \approx (\text{small positive})(\text{negative}) = \text{negative}$. So, $$\lim_{x \to 2^-} f(x) = +\infty, \quad \lim_{x \to 2^+} f(x) = -\infty.$$ - As $x \to 3$, numerator $= 2(3)^2 - 1 = 18 - 1 = 17$. - Denominator $\to 0$, check sign: - For $x \to 3^-$, denominator $= (3^- - 2)(3^- - 3) \approx (\text{positive})(\text{small negative}) = \text{negative}$. - For $x \to 3^+$, denominator $= (3^+ - 2)(3^+ - 3) \approx (\text{positive})(\text{small positive}) = \text{positive}$. So, $$\lim_{x \to 3^-} f(x) = -\infty, \quad \lim_{x \to 3^+} f(x) = +\infty.$$ 5. **Find asymptotes:** - Vertical asymptotes at $x=2$ and $x=3$ because denominator zero and limits infinite. - Horizontal or oblique asymptote: Degree numerator = degree denominator = 2. Leading coefficients ratio: $\frac{2}{1} = 2$. So horizontal asymptote is $y=2$. 6. **First derivative:** Use quotient rule: $$f'(x) = \frac{(4x)(x^2 - 5x + 6) - (2x^2 - 1)(2x - 5)}{(x^2 - 5x + 6)^2}.$$ Simplify numerator: $$N = 4x(x^2 - 5x + 6) - (2x^2 - 1)(2x - 5) = 4x^3 - 20x^2 + 24x - (4x^3 - 10x^2 - 2x + 5) = 4x^3 - 20x^2 + 24x - 4x^3 + 10x^2 + 2x - 5 = (-10x^2) + 26x - 5.$$ So, $$f'(x) = \frac{-10x^2 + 26x - 5}{(x^2 - 5x + 6)^2}.$$ 7. **Inflection points:** Inflection points occur where second derivative changes sign. 8. **Second derivative:** Differentiate $f'(x)$ using quotient rule again or rewrite $f'(x)$ as $\frac{N}{D}$ with $N = -10x^2 + 26x - 5$ and $D = (x^2 - 5x + 6)^2$. Calculate $f''(x) = \frac{N'D - ND'}{D^2}$ where $N' = -20x + 26$, and $D' = 2(x^2 - 5x + 6)(2x - 5)$. 9. **Critical points of $f''(x)$:** Solve $f''(x) = 0$ or undefined to find critical points. 10. **Variation table:** Use sign of $f'(x)$ and $f''(x)$ to determine increasing/decreasing and concavity. 11. **Sketch graph:** Plot vertical asymptotes at $x=2,3$, horizontal asymptote at $y=2$, mark critical points and inflection points, and use variation table to sketch. **Final answers:** - Domain: $\mathbb{R} \setminus \{2,3\}$ - Vertical asymptotes: $x=2$, $x=3$ - Horizontal asymptote: $y=2$ - First derivative: $f'(x) = \frac{-10x^2 + 26x - 5}{(x^2 - 5x + 6)^2}$ - Limits at boundaries: $$\lim_{x \to 2^-} f(x) = +\infty, \quad \lim_{x \to 2^+} f(x) = -\infty,$$ $$\lim_{x \to 3^-} f(x) = -\infty, \quad \lim_{x \to 3^+} f(x) = +\infty.$$