## Precalculus (6th Edition)

domain: $(−\infty,+\infty)$ range: $(−\infty,+\infty)$ Refer to the graph below.
The given line can be graphed using the x and y-intercepts. RECALL: (1) The x-intercept can be found by setting y=0 then solving for x. (2) The y-intercept can be found by setting x=0 then solving for y. Find the x-intercept of the given equation. Set y=0 then solve for x to obtain: \begin{array}{ccc} &3x+2y&=&0 \\&3x+2(0)&=&0 \\&3x&=&=0 \\&\frac{3x}{3}&=&\frac{0}{3} \\&x&=&0\end{array} The x-intercept is $(0,0)$. Find the y-intercept of the given equation. Set x=0 then solve for y to obtain: \begin{array}{ccc} &3x+2y&=&0 \\&3(0)+2y&=&0 \\&2y&=&=0 \\&\frac{2y}{2}&=&\frac{0}{2} \\&y&=&0\end{array} The y-intercept is $(0,0)$. The intercepts are the same so one more point is needed to graph the line. Set $x=2$ then solve for y to obtain: $3x+2y=0 \\3(2)+2y=0 \\6+2y=0 \\6+2y-6=0-6 \\2y=-6 \\\frac{2y}{2}=\frac{-6}{2} \\y=-3$ The line contains the point (2, -3). Graph the line by plotting $(0,0)$ and $(2, -3)$ and connecting them using a line. (Refer to the graph in the answer part above.) The graph covers all x-values therefore the domain is $(−\infty,+\infty)$. The graph covers all y-values therefore the range is $(−\infty,+\infty)$.