## Intermediate Algebra (6th Edition)

$f(x)=\dfrac{1}{4}x-\dfrac{7}{2}$
Using $y=mx+b$ where $m$ is the slope, the given equation, $4x+y=\dfrac{2}{3} ,$ is equivalent to \begin{array}{l}\require{cancel} y=-4x+\dfrac{2}{3} .\end{array} Hence, the slope is $m=-4$. Since perpendicular lines have negative reciprocal slopes, then $m_p=\dfrac{1}{4}$ Using $(2,-3)$ and $m_p= \dfrac{1}{4} ,$ the equation of the line is \begin{array}{l}\require{cancel} y-(-3)=\dfrac{1}{4}(x-2) \\\\ y+3=\dfrac{1}{4}(x-2) \\\\ y+3=\dfrac{1}{4}x-\dfrac{1}{2} \\\\ y=\dfrac{1}{4}x-\dfrac{1}{2}-3 \\\\ y=\dfrac{1}{4}x-\dfrac{1}{2}-\dfrac{6}{2} \\\\ y=\dfrac{1}{4}x-\dfrac{7}{2} .\end{array} In function notation form, this is equivalent to $f(x)=\dfrac{1}{4}x-\dfrac{7}{2} .$