1. **Part A: Find local minima and maxima of f(x)** The function is given as $$f(x) = -5(x - 2)(x - 4)(x + 1)$$ 2. First, expand $f(x)$ to find its derivative more easily: $$f(x) = -5[(x - 2)(x - 4)(x + 1)]$$ Calculate $(x - 2)(x - 4) = x^2 - 6x + 8$ Then, $$f(x) = -5(x^2 - 6x + 8)(x + 1) = -5(x^3 - 5x^2 + 2x + 8)$$ So, $$f(x) = -5x^3 + 25x^2 - 10x - 40$$ 3. Find the first derivative: $$f'(x) = \frac{d}{dx} (-5x^3 + 25x^2 -10x - 40) = -15x^2 + 50x - 10$$ 4. Find critical points by solving $f'(x) = 0$: $$-15x^2 + 50x -10 = 0$$ Divide both sides by -5: $$3x^2 - 10x + 2 = 0$$ Use quadratic formula: $$x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} = \frac{10 \pm \sqrt{100 - 24}}{6} = \frac{10 \pm \sqrt{76}}{6}$$ Simplify $\sqrt{76} = 2\sqrt{19}$: $$x = \frac{10 \pm 2\sqrt{19}}{6} = \frac{5 \pm \sqrt{19}}{3}$$ Approximate values: $$x_1 \approx \frac{5 - 4.3589}{3} = 0.2137$$ $$x_2 \approx \frac{5 + 4.3589}{3} = 3.12$$ 5. Determine nature of critical points using second derivative: $$f''(x) = \frac{d}{dx} (-15x^2 + 50x - 10) = -30x + 50$$ Evaluate at $x_1$: $$f''(0.2137) = -30 \cdot 0.2137 + 50 = -6.411 + 50 = 43.59 > 0$$ so local minimum at $x_1$. Evaluate at $x_2$: $$f''(3.12) = -30 \cdot 3.12 + 50 = -93.6 + 50 = -43.6 < 0$$ so local maximum at $x_2$. 6. Find coordinates of local extrema by plugging critical points into $f(x)$: $$f(0.2137) = -5(0.2137-2)(0.2137-4)(0.2137+1)$$ Calculate: $$0.2137-2 = -1.7863, \quad 0.2137-4 = -3.7863, \quad 0.2137+1 = 1.2137$$ Product: $-1.7863 \times -3.7863 \times 1.2137 \approx 8.208$ Then, $$f(0.2137) = -5 \times 8.208 = -41.04$$ Similarly for maximum: $$f(3.12) = -5(3.12-2)(3.12-4)(3.12+1)$$ Calculate: $$3.12-2=1.12, \quad 3.12-4=-0.88, \quad 3.12+1=4.12$$ Product: $1.12 \times -0.88 \times 4.12 = -4.06$ Then, $$f(3.12) = -5 \times -4.06 = 20.3$$ **Answer Part A:** - Local minimum at approximately $\boxed{(0.21, -41.04)}$ - Local maximum at approximately $\boxed{(3.12, 20.3)}$ --- 7. **Part B: Hole of h at (c, L) and integers a, b** Given: $$h(x) = \frac{f(x)}{g(x)} = \frac{-5(x - 2)(x - 4)(x + 1)}{2(x - 4)(x + 3)}$$ Simplify: $$h(x) = \frac{-5(x - 2)(x + 1)}{2(x + 3)}$$ with $x \neq 4$ (denominator zero at $x=4$) The graph of $h$ has a hole where numerator and denominator share a factor zero, here at $x=c=4$. Find the limit $L$: $$L = \lim_{x\to4} h(x) = \lim_{x\to4} \frac{-5(x-2)(x+1)}{2(x+3)}$$ Evaluate directly: $$L = \frac{-5(4-2)(4+1)}{2(4+3)} = \frac{-5 \times 2 \times 5}{2 \times 7} = \frac{-50}{14} = -\frac{25}{7} \approx -3.5714$$ We know $a < L < b$ with $a,b$ consecutive integers. Since $-4 < -3.5714 < -3$, the integers are $oxed{a = -4, b = -3}$. Reason: The hole coordinate has $x=4$ where original and simplified versions differ, and limit value $L$ is between consecutive integers $a$ and $b$. --- 8. **Part C: Horizontal or slant asymptotes of h** Recall: $$h(x) = \frac{f(x)}{g(x)} = \frac{-5x^3 + 25x^2 - 10x - 40}{2x^2 + 2x - 24}$$ Degree numerator = 3, degree denominator = 2. Since degree numerator $>$ degree denominator by 1, there is no horizontal asymptote but there is a slant (oblique) asymptote. Find slant asymptote by polynomial division: divide numerator by denominator. Divide: $$\frac{-5x^3 + 25x^2 - 10x - 40}{2x^2 + 2x - 24}$$ Step 1: Leading term division: $$-5x^3 \div 2x^2 = -\frac{5}{2}x$$ Multiply divisor by $-\frac{5}{2}x$: $$-\frac{5}{2}x \times (2x^2 + 2x - 24) = -5x^3 - 5x^2 + 60x$$ Subtract: $$(-5x^3 + 25x^2 -10x - 40) - (-5x^3 - 5x^2 + 60x) = 0 + 30x^2 - 70x - 40$$ Step 2: Divide leading term: $$30x^2 \div 2x^2 = 15$$ Multiply divisor by 15: $$15(2x^2 + 2x - 24) = 30x^2 + 30x - 360$$ Subtract: $$(30x^2 -70x -40) - (30x^2 + 30x - 360) = 0 - 100x + 320$$ Remainder: $-100x + 320$ Slant asymptote: $$y = -\frac{5}{2} x + 15$$ **Answer Part C:** - No horizontal asymptotes. - One slant asymptote: $$y = -\frac{5}{2} x + 15$$ with slope $-\frac{5}{2}$. --- **Summary:** - Part A local extrema: minimum at $(0.21,-41.04)$, maximum at $(3.12,20.3)$. - Part B hole at $x=4$ with $L = -\frac{25}{7} \approx -3.5714$, integers $a=-4$, $b=-3$. - Part C slant asymptote: $y = -\frac{5}{2} x + 15$, no horizontal asymptote.