Answer
$f(x)=-\dfrac{1}{3}x+\dfrac{5}{3}$
Work Step by Step
Using $y=mx+b$ where $m$ is the slope, the slope-intercept form of the given equation, $
3x-y=4
$ is
\begin{array}{l}\require{cancel}
-y=-3x+4
\\\\
y=3x-4
.\end{array}
Hence, the slope is $3$.
Since the line is perpendicular to the equation of the line above, use the negative reciprocal slope. Using $y-y_1=m(x-x_1)$ or the Point-Slope form, the equation of the line with slope $-\dfrac{1}{3}$ and passing through $(-1,2)$ is
\begin{array}{l}\require{cancel}
y-2=-\dfrac{1}{3}(x-(-1))
\\\\
y-2=-\dfrac{1}{3}(x+1)
\\\\
y-2=-\dfrac{1}{3}x-\dfrac{1}{3}
\\\\
y=-\dfrac{1}{3}x-\dfrac{1}{3}+2
\\\\
y=-\dfrac{1}{3}x-\dfrac{1}{3}+\dfrac{6}{3}
\\\\
y=-\dfrac{1}{3}x+\dfrac{5}{3}
.\end{array}
In function notation, this is equivalent to $
f(x)=-\dfrac{1}{3}x+\dfrac{5}{3}
.$
