## Precalculus (6th Edition)

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