## Precalculus (6th Edition)

$C$
RECALL: (1) The point-slope form of a line's equation is $y-y-1=m(x-x_1)$ where $m$=slope and $(x_1, y_1)$ is a point on the line. (2) The slope can be interpreted as the ratio of the rise (change in $y$) over run (change in $x$). Thus, the given equation has a slope of $\dfrac{3}{2}$ This means that the rise is $3$ units and the run is $2$ units. Notice for the graph in Option $C$, from the point $(-1, -4)$ to the point $(1, -1)$, there is a rise of $3$ units and a run of $2$ units. Thus, its slope is $\dfrac{3}{2}$. Therefore, the graph of the given equation is the one in Option $C$.