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# How do you graph $y=-\dfrac{3}{5}x+1$ ?

Last updated date: 20th Jun 2024
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Hint: For answering this question we need to draw the graph of $y=-\dfrac{3}{5}x+1$ . For drawing the graph of the straight line we will find the intercepts of the given equation of the straight line and plot them and join them and extend the line formed. The intercepts are the respective points when the other variable is zero.

Complete step-by-step solution:
Now considering from the question we have to draw the graph of $y=-\dfrac{3}{5}x+1$ .
From the basics of the concept we know that for drawing the graph of the straight line we need to find the intercepts of the given equation of the straight line and plot them and join them and extend the line formed.
The intercepts are the respective points when the other variable is zero.
The $\text{x-intercept}$ is given by substituting $y=0$ in the given equation of the straight line.
After substituting we will have
\begin{align} & \Rightarrow 0=-\dfrac{3}{5}x+1 \\ & \Rightarrow -1=-\dfrac{3}{5}x \\ & \Rightarrow x=\dfrac{5}{3} \\ \end{align}
The $\text{y-intercept}$ is given by substituting $x=0$ in the given equation of the straight line.
After substituting we will have
\begin{align} & \Rightarrow y=-\dfrac{3}{5}\left( 0 \right)+1 \\ & \Rightarrow y=1 \\ \end{align}
So now we have two points lying on the straight line $\left( 0,1 \right)$ and $\left( \dfrac{5}{3},0 \right)$ .
Now we will plot these two points and join them and extend them to form a straight line. The graph of the straight line is shown below.

Note: For answering this question we need to draw the graph of the given equation of the straight line so we need to be careful while plotting the points and extending the line. We should be careful while performing the calculations. The slope of this given straight line $y=-\dfrac{3}{5}x+1$ is $\dfrac{-3}{5}$ as we know that the slope intercept form is given as $y=mx+c$ where $m$ is the slope.