## Intermediate Algebra (12th Edition)

$\text{Slope-Intercept Form: } y=-\dfrac{1}{4}x-1 \\\text{Standard Form: } x+4y=-4$
$\bf{\text{Solution Outline:}}$ To find the equation of the line with the following properties: \begin{array}{l}\require{cancel} \text{Slope: } -\dfrac{1}{4} \\\text{$y$-intercept: } (0,-1) ,\end{array} use the Slope-Intercept Form of linear equations. Give the equation in the Slope-Intercept Form and in the Standard Form. $\bf{\text{Solution Details:}}$ With the given properties, then $m=-\dfrac{1}{4}$ and $b=-1 .$ Using $y=mx+b$ (where $m$ is the slope and $b$ is the $y$-intercept) or the Slope-Intercept Form of linear equations, then the equation of the line is \begin{array}{l}\require{cancel} y=mx+b \\\\ y=-\dfrac{1}{4}x+(-1) \\\\ y=-\dfrac{1}{4}x-1 .\end{array} Using the properties of equality, in the form $ax+by=c$ or the Standard Form, the equation above is equivalent to \begin{array}{l}\require{cancel} 4(y)=4\left( -\dfrac{1}{4}x-1 \right) \\\\ 4y=-x-4 \\\\ x+4y=-4 .\end{array} The equation of the line is \begin{array}{l}\require{cancel} \text{Slope-Intercept Form: } y=-\dfrac{1}{4}x-1 \\\text{Standard Form: } x+4y=-4 .\end{array}