Q. 55 - Precalculus Enhanced with Graphing Utilities

Question

Graph the equations given in Example $4$.

Step-by-Step Solution

Answer

The graph for the system of equations $\left\{\begin{array}{l}{x}^2+x+y^2-3y+2=0\\ x+1+\frac{y^2-y}{x}=0\end{array}\right.$ is,

Step 1 The equations in the example 4 are,

$\left\{\begin{array}{l}{x}^2+x+y^2-3y+2=0\\ x+1+\frac{y^2-y}{x}=0\end{array}\right.$

Consider the first equation and simplify it.

$$\begin{gathered} x^2+x+y^2-3y+2=0 \\ {x}^2+x+y^2-3y=-2 \\ {x}^2+x+{\left(\frac{1}{2}\right)}^2+y^2-3y+{\left(\frac{3}{2}\right)}^2=-2+{\left(\frac{1}{2}\right)}^2+{\left(\frac{3}{2}\right)}^2 \\ {\left(x+\frac{1}{2}\right)}^2+{\left(y-\frac{3}{2}\right)}^2=-2+\frac{1}{4}+\frac{9}{4} \\ {\left(x+\frac{1}{2}\right)}^2+{\left(y-\frac{3}{2}\right)}^2=\frac{1}{2} \end{gathered}$$

Step 2 The obtained equation takes the form of x - h 2 + y - k 2 = r 2 , which is the equation of a circle.

So, the circle has radius $\frac{1}{2}$ and the center $\left(-\frac{1}{2},\frac{3}{2}\right)$.

Now consider the second equation and simplify it.

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

So, the second equation also takes the form of the equation of the circle with radius $\frac{1}{2}$ and the center $\left(-\frac{1}{2},\frac{1}{2}\right)$.

Step 3 Now sketch a graph for the given equations.

From the second equation we can see that $x$ cannot be zero. To indicate this, there are holes at points where the circle touches the $y$-axis.