How do you graph $y = \dfrac{8}{3}x - 3$?

Hint:First of all this is a very simple and a very easy problem. The general equation of a straight line is $y = mx + c$, where $m$ is the gradient and $y = c$ is the value where the line cuts the y-axis. The number $c$ is called the intercept on the y-axis. Based on this provided information we try to find the graph of the given straight line.

Complete step by step answer:Consider the given linear equation, as given below:
$ \Rightarrow y = \dfrac{8}{3}x - 3$
Now the given straight line is already in the standard form of the general equation of a straight line.
The slope of the straight line $y = \dfrac{8}{3}x - 3$, on comparing with the straight line $y = mx + c$,
Here the slope is $m$, and here on comparing the coefficients of $x$, as shown below:
$ \Rightarrow m = \dfrac{8}{3}$
So the slope of the given straight line$y = \dfrac{8}{3}x - 3$ is $\dfrac{8}{3}$.
Now finding the intercept of the line $y = \dfrac{8}{3}x - 3$, on comparing with the straight line $y = mx + c$, Here the intercept is $c$, and here on comparing the constants of the straight lines,
$ \Rightarrow c = - 3$
So the intercept of the given straight line$y = \dfrac{8}{3}x - 3$ is -3.
Now plotting the straight line with slope $\dfrac{8}{3}$ and a y-intercept of -3, as shown below, here the y-intercept is negative, whereas the slope is positive.


Note:
Please note that while solving such kind of problems, we should understand that if the y-intercept value is zero, then the straight line is passing through the origin, which is in the equation of $y = mx + c$, if $c = 0$, then the equation becomes $y = mx$, and this line passes through the origin, whether the slope is positive or negative.