1. **Problem:** Find the function and analyze $y = x^2 + 2x + 1$. 2. **Formula and rules:** This is a quadratic function of the form $y = ax^2 + bx + c$ where $a=1$, $b=2$, and $c=1$. 3. **Intermediate work:** Factorize the quadratic: $$y = (x+1)^2$$ 4. **Explanation:** The graph is a parabola opening upwards with vertex at $(-1,0)$. --- 1. **Problem:** Analyze $y = x^3 - 2x^2 + 4$. 2. **Formula and rules:** This is a cubic polynomial. 3. **Intermediate work:** Find critical points by differentiating: $$\frac{dy}{dx} = 3x^2 - 4x$$ Set derivative to zero: $$3x^2 - 4x = 0 \Rightarrow x(3x - 4) = 0 \Rightarrow x=0 \text{ or } x=\frac{4}{3}$$ 4. **Explanation:** These points are candidates for local maxima or minima. --- 1. **Problem:** Given implicit equation $x^3 - 2x^2 y^3 - \frac{5}{4} y^2 = 0$, find $\frac{dy}{dx}$ at $x=1$, $y=1$. 2. **Formula and rules:** Use implicit differentiation: Differentiate both sides w.r.t. $x$: $$3x^2 - 2(2x y^3 + x^2 3y^2 \frac{dy}{dx}) - \frac{5}{4} 2y \frac{dy}{dx} = 0$$ 3. **Intermediate work:** Simplify: $$3x^2 - 4x y^3 - 6x^2 y^2 \frac{dy}{dx} - \frac{5}{2} y \frac{dy}{dx} = 0$$ Group $\frac{dy}{dx}$ terms: $$-6x^2 y^2 \frac{dy}{dx} - \frac{5}{2} y \frac{dy}{dx} = -3x^2 + 4x y^3$$ Factor $\frac{dy}{dx}$: $$\frac{dy}{dx} (-6x^2 y^2 - \frac{5}{2} y) = -3x^2 + 4x y^3$$ Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{-3x^2 + 4x y^3}{-6x^2 y^2 - \frac{5}{2} y}$$ Evaluate at $x=1$, $y=1$: $$\frac{dy}{dx} = \frac{-3(1)^2 + 4(1)(1)^3}{-6(1)^2 (1)^2 - \frac{5}{2} (1)} = \frac{-3 + 4}{-6 - 2.5} = \frac{1}{-8.5} = -\frac{2}{17}$$ 4. **Explanation:** The slope of the curve at $(1,1)$ is $-\frac{2}{17}$. --- 1. **Problem:** Velocity $V = \frac{5t^3 - 3t^2 + 2t + 1}{3!}$, find minimum and maximum velocity. 2. **Formula and rules:** $3! = 6$, so $$V = \frac{5t^3 - 3t^2 + 2t + 1}{6}$$ 3. **Intermediate work:** Differentiate velocity to find critical points: $$\frac{dV}{dt} = \frac{15t^2 - 6t + 2}{6} = \frac{15t^2 - 6t + 2}{6}$$ Set derivative to zero: $$15t^2 - 6t + 2 = 0$$ Discriminant: $$\Delta = (-6)^2 - 4 \times 15 \times 2 = 36 - 120 = -84 < 0$$ No real roots, so no critical points. 4. **Explanation:** Since $\frac{dV}{dt}$ has no real zeros, velocity has no local minima or maxima; it is monotonic. --- 1. **Problem:** Given $x_1 = 2 + t$, $x_2 = -3 + 4t$, and $x_3 = \frac{z_2}{z_1}$. 2. **Formula and rules:** No further info on $z_1$, $z_2$; cannot proceed without definitions. --- 1. **Problem:** Determine $y_3$ in terms of $y_1^2$. 2. **Formula and rules:** Insufficient data to relate $y_3$ and $y_1$. --- 1. **Problem:** Represent $x_1$, $x_2$, $x_3$ on an Argand diagram. 2. **Formula and rules:** $x_1$, $x_2$ appear to be real expressions; $x_3$ depends on $z_1$, $z_2$ (complex numbers?). 3. **Explanation:** Without explicit complex forms, cannot plot. --- 1. **Problem:** Express $x_2$ in terms of $x_1$ and $x_3$. 2. **Formula and rules:** Insufficient data to express $x_2$ in terms of $x_1$ and $x_3$. --- **Summary:** Problems 1, 2, 3, and 4 are fully solved with detailed steps.