1. Problem: Sketch the curve $y = x^3 - 6x^4 + 6$ by finding points of intersection with axes and using max/min points. Step 1: Find x-intercepts by setting $y=0$: $$x^3 - 6x^4 + 6 = 0$$ This is a quartic equation; approximate or factor if possible. Step 2: Find y-intercept by setting $x=0$: $$y = 0 - 0 + 6 = 6$$ So, y-intercept is $(0,6)$. Step 3: Use max/min points from Progress Exercises 6.6 (assumed given). 2. Problem: Sketch the curve $y = 2x - 3x^2$. Step 1: Find x-intercepts by setting $y=0$: $$2x - 3x^2 = 0 \implies x(2 - 3x) = 0 \implies x=0 \text{ or } x=\frac{2}{3}$$ Step 2: Find y-intercept by setting $x=0$: $$y=0$$ Step 3: Find max/min by differentiating: $$y' = 2 - 6x$$ Set $y'=0$: $$2 - 6x = 0 \implies x=\frac{1}{3}$$ Step 4: Evaluate $y$ at $x=\frac{1}{3}$: $$y = 2\times\frac{1}{3} - 3\times\left(\frac{1}{3}\right)^2 = \frac{2}{3} - 3\times\frac{1}{9} = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$$ So, max point at $\left(\frac{1}{3}, \frac{1}{3}\right)$. 3. Problem: Sketch the curve $P = -2Q^3 + 8Q$. Step 1: Find intercepts: Set $P=0$: $$-2Q^3 + 8Q = 0 \implies -2Q(Q^2 - 4) = 0 \implies Q=0, Q=\pm 2$$ Step 2: Find $P$ when $Q=0$: $$P=0$$ Step 3: Use max/min points from Progress Exercises 6.6 (assumed given). 4. Problem: Sketch the curve $C = Y + 100$. Step 1: This is a linear function with slope 1 and y-intercept 100. Step 2: Intercepts: Set $C=0$: $$0 = Y + 100 \implies Y = -100$$ Set $Y=0$: $$C = 100$$ Summary: - For each curve, intercepts and max/min points are identified. - Use these points to sketch the curves accurately.