How do you graph $4x + y = - 8$ using intercepts?

Hint: We have to make the graph of the equation using intercepts. First convert the equation of line in intercept form i.e. $\dfrac{x}{a} + \dfrac{y}{b} = 1$, here $a$ and $b$ are $x$ and $y$ intercepts respectively. Then draw the straight line using the resultant $x$ and $y$ intercepts.

Complete step-by-step solution:
According to the question, a straight line equation is given to us and we have to draw its graph using intercepts.
First we’ll convert the equation in its intercept form. We know that the intercept form of straight-line equation is $\dfrac{x}{a} + \dfrac{y}{b} = 1$, here $a$ and $b$ are $x$ and $y$ intercepts respectively. So for the above equation we have:
$
   \Rightarrow 4x + y = - 8 \\
   \Rightarrow \dfrac{{4x}}{{ - 8}} + \dfrac{y}{{ - 8}} = 1 \\
   \Rightarrow \dfrac{x}{{ - 2}} + \dfrac{y}{{ - 8}} = 1
 $
In comparison we have $a = - 2$ and $b = - 8$. Thus $x$ intercept is -2 and $y$ intercept is -8.
From this, we can say that the straight line will cut both the $x$ and $y$ axes on the negative side and it will pass through the third quadrant while cutting the axes.
Based on these obtained results, the graph can be made as shown below:


Note: If x and y intercepts of a line are known to us, we can easily determine the equation of line. Let $a$ and $b$ are the x and y intercepts respectively of the line, then the equation of line is:
$ \Rightarrow \dfrac{x}{a} + \dfrac{y}{b} = 1$
And if only y intercept is known along with the slope of line then also its equation can be easily determined. If $m$ is the slope and $c$ is the y intercept of the line, then the equation of line is:
$ \Rightarrow y = mx + c$