How do you graph \[y=\dfrac{1}{4}x-5\]?

Hint: In this problem, we have to graph the given equation. To graph the equation, we have to find the x-intercept and the y-intercept. We know that at the x-intercept, the value of y is 0 and at y-intercept, the value of x is 0. By substituting x equal to 0, we can get the y-intercept and y equal to 0, we can get x-intercept and we can plot the points in the graph.

Complete step-by-step solution:
We know that the given equation is,
 \[y=\dfrac{1}{4}x-5\]……. (1)
Now we can find the x-intercept.
We know that at x-intercept, the value of y is 0.
We can substitute for y = 0, in equation (1), we get
\[\begin{align}
  & \Rightarrow 0=\dfrac{1}{4}x-5 \\
 & \Rightarrow \dfrac{1}{4}x=5 \\
\end{align}\]
Now we can multiply by 4 on both sides and cancel similar terms, we get
\[\begin{align}
  & \Rightarrow \dfrac{4x}{4}=5\times 4 \\
 & \Rightarrow x=20 \\
\end{align}\]
The x-intercept is \[\left( 20,0 \right)\]
Now we can find y-intercept.
We know that at y-intercept, the value of x is 0.
We can substitute for x = 0, in equation (1), we get
\[\begin{align}
  & \Rightarrow y=\dfrac{1}{4}\left( 0 \right)-5 \\
 & \Rightarrow y=-5 \\
\end{align}\]
The y-intercept is \[\left( 0,-5 \right)\] .
We can also find other points, through which the line passes.
For y = -2, the value of x is
\[\begin{align}
  & \Rightarrow -2=\dfrac{x}{4}-5 \\
 & \Rightarrow \dfrac{x}{4}=3 \\
 & \Rightarrow x=12 \\
\end{align}\]
Therefore, the x-intercept is \[\left( 20,0 \right)\], the y-intercept is \[\left( 0,-5 \right)\] and the other point is \[\left( 12,-2 \right)\]
Now, we can plot the graph using the x-intercept \[\left( 20,0 \right)\], the y-intercept \[\left( 0,-5 \right)\]and the other point \[\left( 12,-2 \right)\].


Note: Students make mistakes while finding the value for x-intercept and the y-intercept, we should know that at the x-intercept, the value of y is 0 and at y-intercept, the value of x is 0. By substituting x equal to 0, we can get the y-intercept and y equal to 0, we can get x-intercept and we can plot the points in the graph.