4.5 Use the slope–intercept form of an equation of a line (Page 2/12)

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Identify the slope and y -intercept of the line $y = \frac{2}{5} x - 1$ .

$\frac{2}{5}; (0, −1)$

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Identify the slope and y -intercept of the line $y = - \frac{4}{3} x + 1$ .

$- \frac{4}{3}; (0, 1)$

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When an equation of a line is not given in slope–intercept form, our first step will be to solve the equation for $y$ .

Identify the slope and y -intercept of the line with equation $x + 2 y = 6$ .

Solution

This equation is not in slope–intercept form. In order to compare it to the slope–intercept form we must first solve the equation for $y$ .

Solve for y .	$x + 2 y = 6$
Subtract x from each side.
Divide both sides by 2.
Simplify.
$(Remember: \frac{a + b}{c} = \frac{a}{c} + \frac{b}{c})$
Simplify.
Write the slope–intercept form of the equation of the line.
Write the equation of the line.
Identify the slope.
Identify the y -intercept.

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Identify the slope and y -intercept of the line $x + 4 y = 8$ .

$- \frac{1}{4}; (0, 2)$

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Identify the slope and y -intercept of the line $3 x + 2 y = 12$ .

$- \frac{3}{2}; (0, 6)$

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Graph a line using its slope and intercept

Now that we know how to find the slope and y -intercept of a line from its equation, we can graph the line by plotting the y -intercept and then using the slope to find another point.

How to graph a line using its slope and intercept

Graph the line of the equation $y = 4 x - 2$ using its slope and y -intercept.

Solution

The figure shows the steps to graph the equation y equals 4x minus 2. Step 1 is to find the slope intercept form of the equation. The equation is already in slope intercept form.

Step 2 is to identify the slope and y-intercept. Use the equation y equals m x, plus b. The equation y equals m x, plus b is shown with the variable m colored red and the variable b colored blue. Below that is the equation y equals 4 x, plus -2. The number 4 is colored red and -2 is colored blue. From this equation we can see that m equals 4 and b equals -2 so the slope is 4 and the y-intercept is the point (0, negative 2).

Step 3 is to plot the y-intercept. An x y-coordinate plane is shown with the x-axis of the plane running from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. The point (0, negative 2) is plotted.

Step 4 is to use the slope formula m equals rise over run to identify the rise and the run. Since m equals 4, rise over run equals 4 over 1. From this we can determine that the rise is 4 and the run is 1.

Step 5 is to start at they-intercept, count out the rise and run to mark the second point. So start at the point (0, negative 2) and count the rise and the run. The rise is up 4 and the run is right 1. On the x y-coordinate plane is a red vertical line starts at the point (0, negative 2) and rises 4 units at its end a red horizontal line runs 1 unit to end at the point (1, 2). The point (1, 2) is plotted.

Step 6 is to connect the points with a line. On the x y-coordinate plane the points (0, negative 2) and (1, 2) are plotted and a line runs through the two points. The line is the graph of y equals 4 x, minus 2.

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Graph the line of the equation $y = 4 x + 1$ using its slope and y -intercept.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The points (0, 1) and (1, 5) are plotted on the line.

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Graph the line of the equation $y = 2 x - 3$ using its slope and y -intercept.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The points (0, negative 3) and (1, negative 1) are plotted on the line.

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Graph a line using its slope and y -intercept.

Find the slope-intercept form of the equation of the line.
Identify the slope and y -intercept.
Plot the y -intercept.
Use the slope formula $m = \frac{rise}{run}$ to identify the rise and the run.
Starting at the y -intercept, count out the rise and run to mark the second point.
Connect the points with a line.

Graph the line of the equation $y = − x + 4$ using its slope and y -intercept.

Solution

	$y = m x + b$
The equation is in slope–intercept form.	$y = − x + 4$
Identify the slope and y -intercept.	$m = −1$
	y -intercept is (0, 4)
Plot the y -intercept.	See graph below.
Identify the rise and the run.	$m = \frac{−1}{1}$
Count out the rise and run to mark the second point.	rise −1, run 1
Draw the line.
To check your work, you can find another point on the line and make sure it is a solution of the equation. In the graph we see the line goes through (4, 0).
Check. $\begin{matrix} y & = & − x + 4 \\ 0 & \overset{?}{=} & - 4 + 4 \\ 0 & = & 0 ✓ \end{matrix}$

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Graph the line of the equation $y = − x - 3$ using its slope and y -intercept.

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Graph the line of the equation $y = − x - 1$ using its slope and y -intercept.

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Graph the line of the equation $y = - \frac{2}{3} x - 3$ using its slope and y -intercept.

Solution

	$y = m x + b$
The equation is in slope–intercept form.	$y = - \frac{2}{3} x - 3$
Identify the slope and y -intercept.	$m = - \frac{2}{3}$ ; y -intercept is (0, −3)
Plot the y -intercept.	See graph below.
Identify the rise and the run.
Count out the rise and run to mark the second point.
Draw the line.

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Graph the line of the equation $y = - \frac{5}{2} x + 1$ using its slope and y -intercept.

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Graph the line of the equation $y = - \frac{3}{4} x - 2$ using its slope and y -intercept.

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Graph the line of the equation $4 x - 3 y = 12$ using its slope and y -intercept.

Solution

	$4 x - 3 y = 12$
Find the slope–intercept form of the equation.	$−3 y = −4 x + 12$
	$- \frac{3 y}{3} = \frac{−4 x + 12}{−3}$
The equation is now in slope–intercept form.	$y = \frac{4}{3} x - 4$
Identify the slope and y -intercept.	$m = \frac{4}{3}$
	y -intercept is (0, −4)
Plot the y -intercept.	See graph below.
Identify the rise and the run; count out the rise and run to mark the second point.
Draw the line.

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Source: OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2

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