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Question

Question: How do you graph \(y = \dfrac{8}{3}x - 3\)?...

How do you graph y=83x3y = \dfrac{8}{3}x - 3?

Explanation

Solution

First of all this is a very simple and a very easy problem. The general equation of a straight line is y=mx+cy = mx + c, where mm is the gradient and y=cy = c is the value where the line cuts the y-axis. The number cc is called the intercept on the y-axis. Based on this provided information we try to find the graph of the given straight line.

Complete step by step answer: Consider the given linear equation, as given below:
y=83x3\Rightarrow y = \dfrac{8}{3}x - 3
Now the given straight line is already in the standard form of the general equation of a straight line.
The slope of the straight line y=83x3y = \dfrac{8}{3}x - 3, on comparing with the straight line y=mx+cy = mx + c,
Here the slope is mm, and here on comparing the coefficients of xx, as shown below:
m=83\Rightarrow m = \dfrac{8}{3}
So the slope of the given straight liney=83x3y = \dfrac{8}{3}x - 3 is 83\dfrac{8}{3}.
Now finding the intercept of the line y=83x3y = \dfrac{8}{3}x - 3, on comparing with the straight line y=mx+cy = mx + c, Here the intercept is cc, and here on comparing the constants of the straight lines,
c=3\Rightarrow c = - 3
So the intercept of the given straight liney=83x3y = \dfrac{8}{3}x - 3 is -3.
Now plotting the straight line with slope 83\dfrac{8}{3} and a y-intercept of -3, as shown below, here the y-intercept is negative, whereas the slope is positive.

Note:
Please note that while solving such kind of problems, we should understand that if the y-intercept value is zero, then the straight line is passing through the origin, which is in the equation of y=mx+cy = mx + c, if c=0c = 0, then the equation becomes y=mxy = mx, and this line passes through the origin, whether the slope is positive or negative.