Question

Consider the graph of the solutions of the equation$y^3+xy+2=0$

(a) Find all points on the graph with an x-coordinate of x = 1, and then find the slope of the tangent line at each of these points.

(b) Find all points on the graph with a y-coordinate of y = 1, and then find the slope of the tangent line at each of these points.

(c) Find all points where the graph has a horizontal tangent line.

(d) Find all points where the graph has a vertical tangent line.

Short answer

Part (a): $$\begin{gathered} \left(1,-2\right)\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{graph}. \\ \frac{d\mathrm{y}}{d\mathrm{x}}=\frac{2}{13} \end{gathered}$$

Part (b): $$\begin{gathered} \left(-3,1\right)\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{graph}. \\ \frac{d\mathrm{y}}{d\mathrm{x}}=-\infty \end{gathered}$$

Part (c): $\mathrm{There}\mathrm{is}\mathrm{no}\mathrm{point}\mathrm{where}\mathrm{tangent}\mathrm{line}\mathrm{is}\mathrm{horizontal}.$

Part (d): $\mathrm{There}\mathrm{is}\mathrm{no}\mathrm{point}\mathrm{where}\mathrm{tangent}\mathrm{line}\mathrm{is}\mathrm{vertical}.$

Step-by-step solution

Step 1. Given information is:

$y^3+xy+2=0$

Part (a) Step 1. Calculating dy/dx

$$\begin{gathered} \mathrm{Here}, \\ {y}^3+xy+2=0 \\ \frac{d\left({\mathrm{y}}^3+\mathrm{xy}+2\right)}{d\mathrm{x}}=0(\mathrm{Differentiating}\mathrm{both}\mathrm{sides}) \\ 3{\mathrm{y}}^2\frac{d\mathrm{y}}{d\mathrm{x}}+\mathrm{x}\frac{d\mathrm{y}}{d\mathrm{x}}+\mathrm{y}\frac{d\mathrm{x}}{d\mathrm{x}}+0=0 \\ 3{\mathrm{y}}^2\frac{d\mathrm{y}}{d\mathrm{x}}+\mathrm{x}\frac{d\mathrm{y}}{d\mathrm{x}}+\mathrm{y}=0 \\ (3{\mathrm{y}}^2+\mathrm{x})\frac{d\mathrm{y}}{d\mathrm{x}}+\mathrm{y}=0 \\ \frac{d\mathrm{y}}{d\mathrm{x}}=-\frac{y}{3{\mathrm{y}}^2+\mathrm{x}} \end{gathered}$$

Part (a) Step 2. Calculating y and slope 

$$\begin{gathered} \mathrm{Substituting}\mathrm{x}=1\mathrm{into}\mathrm{the}\mathrm{equation}{\mathrm{y}}^3+\mathrm{xy}+2=0: \\ {\mathrm{y}}^3+\left(1\right)\mathrm{y}+2=0 \\ {\mathrm{y}}^3+\mathrm{y}=-2 \\ \mathrm{y}({\mathrm{y}}^2+1)=-2 \\ \mathrm{y}=-2 \\ \mathrm{Thus},\left(1,-2\right)\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{graph}. \\ \mathrm{Thus}\mathrm{the}\mathrm{slope}\mathrm{is}: \\ \mathrm{x}=1,\mathrm{y}=-2 \\ \frac{d\mathrm{y}}{d\mathrm{x}}=-\frac{(-2)}{3{(-2)}^2+1} \\ \frac{d\mathrm{y}}{d\mathrm{x}}=\frac{2}{3\left(4\right)+1} \\ \frac{d\mathrm{y}}{d\mathrm{x}}=\frac{2}{13} \end{gathered}$$

Part (b) Step 1. Calculating x and slope

$$\begin{gathered} \mathrm{Substituting}\mathrm{y}=1\mathrm{into}\mathrm{the}\mathrm{equation}{\mathrm{y}}^3+\mathrm{xy}+2=0: \\ {1}^3+\mathrm{x}+2=0 \\ \mathrm{x}=-3 \\ \mathrm{Thus},\left(-3,1\right)\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{graph}. \\ \mathrm{Thus}\mathrm{the}\mathrm{slope}\mathrm{is}: \\ \mathrm{when}\mathrm{x}=-3,\mathrm{y}=1 \\ \frac{d\mathrm{y}}{d\mathrm{x}}=-\frac{1}{3{\left(1\right)}^2+(-3)} \\ \frac{d\mathrm{y}}{d\mathrm{x}}=\frac{-1}{0}=-\infty \end{gathered}$$

Part (c) Step 1. Finding points

$$\begin{gathered} \mathrm{Substituting}\mathrm{x}=0\mathrm{into}\mathrm{the}\mathrm{equation}{\mathrm{y}}^3+\mathrm{xy}+2=0\mathrm{for}\mathrm{the}\mathrm{tangent}\mathrm{line}\mathrm{to}\mathrm{be}\mathrm{horizontal}: \\ {\mathrm{y}}^3+\left(0\right)\mathrm{y}+2=0 \\ {\mathrm{y}}^3=-2 \\ \mathrm{Thus},\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{point}\mathrm{where}\mathrm{tangent}\mathrm{line}\mathrm{is}\mathrm{horizontal}. \end{gathered}$$

Part (d) Step 1. Finding points

$$\begin{gathered} \mathrm{Substituting}\mathrm{y}=0\mathrm{into}\mathrm{the}\mathrm{equation}{\mathrm{y}}^3+\mathrm{xy}+2=0\mathrm{for}\mathrm{the}\mathrm{tangent}\mathrm{line}\mathrm{to}\mathrm{be}\mathrm{vertical}: \\ {0}^3+\mathrm{x}\left(0\right)+2=0 \\ 2=0 \\ \mathrm{Thus},\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{point}\mathrm{where}\mathrm{tangent}\mathrm{line}\mathrm{is}\mathrm{vertical}. \end{gathered}$$