Answer
The slope-intercept form of the line is $f\left( x \right)=-\frac{2}{3}x-2$.
Work Step by Step
Consider the equation $3x-2y-4=0$.
Isolate the y terms on one side.
$\begin{align}
& 3x-2y-4=0 \\
& -2y=-3x+4
\end{align}$
Divide both sides of the equation by $-2$ to remove the fractional part.
$\begin{align}
& -\frac{2}{-2}y=-\frac{3}{-2}x+\frac{4}{-2} \\
& y=\frac{3}{2}x-2
\end{align}$
Hence, the slope ${{m}_{1}}$ of the line $3x-2y-4=0$ is ${{m}_{1}}=\frac{3}{2}$ and the y-intercept is $-2$.
Now, let the slope of the line $f$ be ${{m}_{2}}$.
So,
$\begin{align}
& {{m}_{1}}\cdot {{m}_{2}}=-1 \\
& \frac{3}{2}\cdot {{m}_{2}}=-1 \\
& {{m}_{2}}=-1\left( \frac{2}{3} \right) \\
& {{m}_{2}}=-\frac{2}{3}
\end{align}$
The y-intercept is equal, that is: $\left( 0,-2 \right)$.
Now, the equation of $f$ having point $\left( 0,-2 \right)$ and slope $-\frac{2}{3}$ is:
$\begin{align}
& f\left( x \right)=mx+b \\
& =-\frac{2}{3}\left( x \right)+\left( -2 \right) \\
& =-\frac{2}{3}x-2
\end{align}$
Hence, the equation of the line in slope-intercept form of the line which is perpendicular to the line $3x-2y-4=0$ and has the same y-intercept is $f\left( x \right)=-\frac{2}{3}x-2$.
