1. **State the problem:** Derive the equation $x^2 + 3x + 4 = 2y^2$ implicitly to find $\frac{dy}{dx}$ and then find the equation of the normal line at the point $(1, 2)$. Also, compute the second derivative $\frac{d^2y}{dx^2}$ at $x=2$ and determine the concavity of the curve locally. 2. **Implicit differentiation:** Start with the equation: $$x^2 + 3x + 4 = 2y^2$$ Differentiate both sides with respect to $x$: $$\frac{d}{dx}(x^2) + \frac{d}{dx}(3x) + \frac{d}{dx}(4) = \frac{d}{dx}(2y^2)$$ $$2x + 3 + 0 = 2 \cdot 2y \frac{dy}{dx}$$ Simplify: $$2x + 3 = 4y \frac{dy}{dx}$$ Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{2x + 3}{4y}$$ 3. **Find the derivative at $(1,2)$:** $$\frac{dy}{dx}\bigg|_{(1,2)} = \frac{2(1) + 3}{4(2)} = \frac{2 + 3}{8} = \frac{5}{8}$$ 4. **Equation of the normal line:** The slope of the tangent at $(1, 2)$ is $\frac{5}{8}$, so the slope of the normal line is the negative reciprocal: $$m_{normal} = -\frac{8}{5}$$ Using point-slope form: $$y - 2 = -\frac{8}{5}(x - 1)$$ Simplify if desired: $$y = 2 - \frac{8}{5}x + \frac{8}{5} = -\frac{8}{5}x + \frac{18}{5}$$ 5. **Compute $\frac{d^2y}{dx^2}$:** Recall: $$\frac{dy}{dx} = \frac{2x + 3}{4y}$$ Rewrite as: $$y' = \frac{2x + 3}{4y}$$ Use the quotient rule or implicit differentiation for $y''$: $$y'' = \frac{d}{dx} \left( \frac{2x + 3}{4y} \right) = \frac{(2)(4y) - (2x + 3)(4y')}{(4y)^2}$$ Substitute $y' = \frac{2x + 3}{4y}$: $$y'' = \frac{8y - 4(2x+3) \frac{2x+3}{4y}}{16 y^2} = \frac{8y - \frac{4(2x+3)^2}{4y}}{16 y^2} = \frac{8y - \frac{(2x+3)^2}{y}}{16 y^2}$$ Multiply numerator terms by $y$ to combine: $$y'' = \frac{8y^2 - (2x + 3)^2}{16 y^3}$$ 6. **Evaluate $y''$ at $x=2$:** Find corresponding $y$ at $x=2$ from original equation: $$2^2 + 3(2) + 4 = 2y^2 \Rightarrow 4 + 6 + 4 = 2y^2 \Rightarrow 14 = 2y^2 \Rightarrow y^2 = 7 \Rightarrow y = \sqrt{7}$$ Calculate numerator: $$8 y^2 - (2x+3)^2 = 8 \cdot 7 - (2(2) + 3)^2 = 56 - (4+3)^2 = 56 - 49 = 7$$ Calculate denominator: $$16 y^3 = 16 (\sqrt{7})^3 = 16 \cdot 7 \sqrt{7} = 112 \sqrt{7}$$ Therefore: $$y''(2) = \frac{7}{112 \sqrt{7}} = \frac{1}{16 \sqrt{7}} > 0$$ 7. **Conclusion:** Since $y''(2) > 0$, the curve is **concave up** locally at $x=2$. **Final answers:** - Normal line at $(1, 2)$: $$y = -\frac{8}{5}x + \frac{18}{5}$$ - Second derivative at $x=2$: $$y'' = \frac{1}{16 \sqrt{7}} > 0$$ (concave up)