Question

How do you graph the equation \[y=\dfrac{1}{3}x-4\]?

Answer 1

Hint: We are asked to draw the graph of the equation \[y=\dfrac{1}{3}x-4\]. The degree of an equation is the highest power of the variable present in it. So, for this equation, the highest power present \[x\] is 1, the degree is also 1. From this, it can be said that this is a linear equation. The graph of a linear equation represents a straight line.

Complete step by step answer:
The general equation of a straight line is \[ax+by+c=0\], where \[a,b,c\] are any real numbers. The given equation is \[y=\dfrac{1}{3}x-4\], the equation can also be written as \[\dfrac{1}{3}x-y-4=0\], comparing with the general equation of straight line, we get \[a=\dfrac{1}{3},b=-1\And c=-4\].
To plot the graph of an equation of the straight line, we should know at least two points, through which the line passes.
To make things simple, let’s take the X-intercept and Y-intercept as the two points. X-intercept is the point where the line crosses X-axis, this means that the Y-coordinate will be \[0\], similarly Y-intercept is the point where the line crosses Y-axis, so X-coordinate will be \[0\]. We will use this property now.
We substitute \[y=0\] in the equation \[\dfrac{1}{3}x-y-4=0\], we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{3}x-0-4=0 \\
 & \Rightarrow \dfrac{1}{3}x-4=0 \\
\end{align}\]
Adding 4 to both sides of the equation we get,
\[\begin{align}
  & \Rightarrow \dfrac{1}{3}x-4+4=0+4 \\
 & \Rightarrow \dfrac{1}{3}x=4 \\
\end{align}\]
We multiply 3 to both sides we get,
\[\begin{align}
  & \Rightarrow \left( \dfrac{1}{3}x \right)3=\left( 4 \right)3 \\
 & \therefore x=12 \\
\end{align}\]
So, the coordinates of the X-intercept are \[\left( 12,0 \right)\].
Similarly, now we substitute \[x=0\] in the equation, we get
\[\Rightarrow \dfrac{1}{3}(0)-y-4=0\]
\[\Rightarrow -y-4=0\]
Adding \[y\] to both sides of the equation, we get
\[\begin{align}
  & \Rightarrow -y-4+y=0+y \\
 & \therefore y=-4 \\
\end{align}\]
So, the coordinates of the Y-intercept are \[\left( 0,-4 \right)\].
 Using these two points we can plot the graph of the equation as follows:

Note: Here, we found the two points which are X-intercept and Y-intercept by substituting either x or \[y\] to be zero, one at a time. We can also find these values by converting the straight-line equation to the equation in intercept form which is, \[\dfrac{x}{a}+\dfrac{y}{b}=1\]. Here, \[a\And b\] are X-intercept and Y-intercept respectively.