Question

How do you graph the linear function \[f(x)=\dfrac{2}{3}x+1\]?

Answer 1

Hint: We are asked to draw the graph of the equation \[f(x)=\dfrac{2}{3}x+1\]. We know that the graph of a linear function is a straight line. To draw the graph of the straight line we should know at least two points through which the line passes.

Complete answer:
The given linear function is \[f(x)=\dfrac{2}{3}x+1\], to make the function look simpler we substitute \[f(x)=y\] in it. So, the function becomes \[y=\dfrac{2}{3}x+1\]. This can also be written as \[\dfrac{2}{3}x-y+1=0\].
The general equation of a straight line is \[ax+by+c=0\], where \[a,b,c\] are any real numbers. The equation we have is \[\dfrac{2}{3}x-y+1=0\], comparing with the general equation of the straight line, we get \[a=\dfrac{2}{3},b=-1\And c=1\].
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 \[\dfrac{2}{3}x-y+1=0\], we get
\[\begin{align}
  & \Rightarrow \dfrac{2}{3}x-0+1=0 \\
 & \Rightarrow \dfrac{2}{3}x+1=0 \\
\end{align}\]
Subtracting 1 from both sides of the equation we get,
\[\begin{align}
  & \Rightarrow \dfrac{2}{3}x+1-1=0-1 \\
 & \Rightarrow \dfrac{2}{3}x=-1 \\
\end{align}\]
We multiply \[\dfrac{3}{2}\] to both sides we get,
\[\begin{align}
  & \Rightarrow \left( \dfrac{2}{3}x \right)\dfrac{3}{2}=\left( -1 \right)\dfrac{3}{2} \\
 & \therefore x=\dfrac{-3}{2} \\
\end{align}\]
So, the coordinates of the X-intercept are \[\left( \dfrac{-3}{2},0 \right)\].
Similarly, now we substitute \[x=0\] in the equation \[\dfrac{2}{3}x-y+1=0\], we get
\[\Rightarrow \dfrac{2}{3}\left( 0 \right)-y+1=0\]
Adding \[y\] to both sides of the equation, we get
\[\begin{align}
 & \Rightarrow -y+1+y=0+y \\
 & \therefore y=1 \\
\end{align}\]
So, the coordinates of the Y-intercept are \[\left( 0,1 \right)\].
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-or \[y\], 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.