How do you graph a line using slope intercept form?

Hint: In this problem, we have to graph a line using slope intercept form. We can first take any of the line equations to graph it using slope intercept form. We know that the slope intercept form equation is \[y=mx+c\] where m is the slope and c is the y-intercept. We can then find the slope and the y-intercept by comparing the example line equation and the slope intercept form, we can also find other points to be plotted in the graph and we can plot the points in the graph.

Complete step-by-step solution:
We know that we can graph a line using the slope intercept form, by finding the slope value and the intercepts value.
We can now take an example line equation,
\[y=5x-9\] ……… (1)
We also know that the general form of the slope intercept form is,
\[y=mx+c\] ……… (2)
Where, m is the slope and c, y-intercept.
Now we can compare equation (1) and (2), we get
Slope, m = 5 and y-intercept, c = -9
We also know that at y-intercept, x = 0.
Therefore, the point at y-intercept is \[\left( 0,-9 \right)\].
Now, we have to find the x-intercept.
We know that at x-intercept, y = 0. Substituting the value of y in equation (1), we get
\[\begin{align}
  & \Rightarrow 0=5x-9 \\
 & \Rightarrow x=\dfrac{9}{5}=1.8 \\
\end{align}\]
Therefore, the point at x-intercept is \[\left( 1.8,0 \right)\].
We can also find some other points to be plotted in the graph where the line passes through.
We can assume for x = 1, then from (1).
\[\begin{align}
  & \Rightarrow y=5\left( 1 \right)-9 \\
 & \Rightarrow y=-4 \\
\end{align}\]
Therefore, the other point is \[\left( 1,-4 \right)\].
Now we can plot the graph using x-intercept \[\left( 1.8,0 \right)\], y-intercept \[\left( 0,-9 \right)\] and the other point \[\left( 1,-4 \right)\].


Note: Students make mistakes in finding the value of x-intercept and y-intercept, we should know that at x-intercept the value of y is 0 and at y-intercept the value of x is 0. We should also concentrate on formulae like slope intercept formulas, to solve these types of problems.