## Intermediate Algebra (12th Edition)

$x\text{-intercept: } (3,0) \\y\text{-intercept: } (0,-4)$
$\bf{\text{Solution Outline:}}$ Use the definition of intercepts to find the $x$- and $y$-intercepts of the given equation, $4x-3y=12 .$ Then plot the points corresponding to the intercepts and connect these with a line to get the graph. $\bf{\text{Solution Details:}}$ The $x$-intercept is the value of $x$ when $y=0.$ Substituting $y=0$ in the given equation, then \begin{array}{l}\require{cancel} 4x-3y=12 \\\\ 4x-3(0)=12 \\\\ 4x-0=12 \\\\ 4x=12 \\\\ x=\dfrac{12}{4} \\\\ x=3 .\end{array} Hence, the $x$-intercept is $(3,0) .$ The $y$-intercept is the value of $y$ when $x=0.$ Substituting $x=0$ in the given equation, then \begin{array}{l}\require{cancel} 4x-3y=12 \\\\ 4(0)-3y=12 \\\\ 0-3y=12 \\\\ -3y=12 \\\\ y=\dfrac{12}{-3} \\\\ y=-4 .\end{array} Hence, the $y$-intercept is $(0,-4) .$ Connecting the following intercepts, \begin{array}{l}\require{cancel} x\text{-intercept: } (3,0) \\y\text{-intercept: } (0,-4) ,\end{array} gives the graph of the given equation.