## Precalculus (6th Edition)

domain: is $(-\infty, +\infty)$ range: $(-\infty, +\infty)$
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} \\&3y-4x&=&0 \\&3(0)-4x&=&0 \\&-4x&=&0 \\&\frac{-4x}{-4}&=&\frac{0}{-4} \\&x&=&0 \end{array} The x-intercept is $(5, 0)$. Find the x-intercept of the given equation. Set $x=0$ then solve for $y$ to obtain: \begin{array}{ccc} \\&3y-4x&=&0 \\&3y-4(0)&=&0 \\&3y&=&0 \\&\frac{3y}{3}&=&\frac{0}{3} \\&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 $x=3$ then solve for $y$ to obtain: $3y-4x=0 \\3y-4(3)=0 \\3y-12=0 \\3y=12 \\\frac{3y}{3}=\frac{12}{3} \\y=4$ The line contains the point $(3, 4)$. Graph the line by plotting $(0, 0)$ and $(3, 4)$ 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 $(-\infty, +\infty)$.