Question

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

Answer 1

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

& \Rightarrow \dfrac{2}{3}x-0+1=0 \\\ & \Rightarrow \dfrac{2}{3}x+1=0 \\\ \end{aligned}$$ Subtracting 1 from both sides of the equation we get, $$\begin{aligned} & \Rightarrow \dfrac{2}{3}x+1-1=0-1 \\\ & \Rightarrow \dfrac{2}{3}x=-1 \\\ \end{aligned}$$ We multiply $$\dfrac{3}{2}$$ to both sides we get, $$\begin{aligned} & \Rightarrow \left( \dfrac{2}{3}x \right)\dfrac{3}{2}=\left( -1 \right)\dfrac{3}{2} \\\ & \therefore x=\dfrac{-3}{2} \\\ \end{aligned}$$ So, the coordinates of the X-intercept are $$\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{aligned} & \Rightarrow -y+1+y=0+y \\\ & \therefore y=1 \\\ \end{aligned}$$ So, the coordinates of the Y-intercept are $$\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.