1. **Problem statement:** Given the function $$f(x) = \frac{x^2 - x + 2}{1} = x^2 - x + 2,$$ we will find its domain, parity, intercepts, limits and asymptotes, first and second derivatives with their sign tables, and sketch the graph. 2. **Domain:** Since the function is a polynomial, it is defined for all real numbers. \[ \text{Domain} = \mathbb{R} \] 3. **Parity:** A function is even if $$f(-x) = f(x)$$ and odd if $$f(-x) = -f(x).$$ Calculate: $$f(-x) = (-x)^2 - (-x) + 2 = x^2 + x + 2 \neq f(x) \text{ and } \neq -f(x)$$ So, the function is neither even nor odd. 4. **Intercepts:** - **y-intercept:** Evaluate at $$x=0$$: $$f(0) = 0^2 - 0 + 2 = 2$$ - **x-intercepts:** Solve $$f(x) = 0$$: $$x^2 - x + 2 = 0$$ Discriminant $$\Delta = (-1)^2 - 4 \times 1 \times 2 = 1 - 8 = -7 < 0,$$ so no real roots. No x-intercepts. 5. **Limits and asymptotes:** As $$x \to \pm \infty,$$ $$f(x) = x^2 - x + 2 \to \pm \infty$$ (dominant term $$x^2$$). No horizontal or oblique asymptotes since degree numerator > denominator. 6. **First derivative:** $$f'(x) = \frac{d}{dx}(x^2 - x + 2) = 2x - 1$$ - **Sign table:** Set $$f'(x) = 0$$: $$2x - 1 = 0 \Rightarrow x = \frac{1}{2}$$ - For $$x < \frac{1}{2}, f'(x) < 0$$ (decreasing) - For $$x > \frac{1}{2}, f'(x) > 0$$ (increasing) 7. **Second derivative:** $$f''(x) = \frac{d}{dx}(2x - 1) = 2$$ - Constant positive, so the function is concave up everywhere. - **Sign table:** always positive. 8. **Summary:** - Domain: all real numbers - Neither even nor odd - y-intercept at (0, 2), no x-intercepts - Limits: $$f(x) \to +\infty$$ as $$x \to \pm \infty$$ - No asymptotes - Increasing on $$(\frac{1}{2}, \infty)$$, decreasing on $$(-\infty, \frac{1}{2})$$ - Concave up everywhere - Minimum at $$x = \frac{1}{2}$$ with value: $$f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} + 2 = \frac{1}{4} - \frac{1}{2} + 2 = \frac{7}{4} = 1.75$$