Q. 56 - Precalculus Enhanced with Graphing Utilities

Question

Graph the equations given in Problem $49$.

Step-by-Step Solution

Answer

The solution for the graph $\left\{\begin{array}{l}{y}^2+y+x^2-x-2=0\\ y+1+\frac{x-2}{y}=0\end{array}\right.$ is,

Step 1 The equations in problem 49 is,

$\left\{\begin{array}{l}{y}^2+y+x^2-x-2=0\\ y+1+\frac{x-2}{y}=0\end{array}\right.$

Consider the first equation and solve it for $y$.

$$\begin{gathered} y^2+y+x^2-x-2=0 \\ {y}^2+y+{\left(\frac{1}{2}\right)}^2+x^2-x+{\left(\frac{1}{2}\right)}^2=2+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2 \\ {\left(y+\frac{1}{2}\right)}^2+{\left(x-\frac{1}{2}\right)}^2=2+\frac{1}{2} \\ {\left(y+\frac{1}{2}\right)}^2+{\left(x-\frac{1}{2}\right)}^2=\frac{5}{2} \\ {\left(y+\frac{1}{2}\right)}^2=-{\left(x-\frac{1}{2}\right)}^2+\frac{5}{2} \\ y+\frac{1}{2}=\pm \sqrt{-{\left(x-\frac{1}{2}\right)}^2+\frac{5}{2}} \\ y=\pm \sqrt{-{\left(x-\frac{1}{2}\right)}^2+\frac{5}{2}}-\frac{1}{2} \end{gathered}$$

Step 2 Now consider the second equation and solve it for y .

$$\begin{gathered} y+1+\frac{x-2}{y}=0 \\ {y}^2+y+x-2=0 \\ {y}^2+y+{\left(\frac{1}{2}\right)}^2=-x+2+{\left(\frac{1}{2}\right)}^2 \\ {\left(y+\frac{1}{2}\right)}^2=-x+2+\frac{1}{4} \\ {\left(y+\frac{1}{2}\right)}^2=-x+\frac{9}{4} \\ y+\frac{1}{2}=\pm \sqrt{-x+\frac{9}{4}} \\ y=\pm \sqrt{-x+\frac{9}{4}}-\frac{1}{2} \end{gathered}$$

Step 3 Now sketch a graph by using the given equations.

From the second equation, we can see that $x\ne 0$. So, the graph has a hole at the points $\left(0,1\right)$ and $\left(0,-2\right)$. The value $x=0$ is extraneous and we discard it.