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Section4.9Summary of Graphing Lines

The previous several sections have demonstrated several methods for plotting a graph of a linear equation. In this section, we review these methods.

Figure4.9.1Alternative Video Lesson

We have learned three forms to write a linear equation:

  • slope-intercept form

    \(y=mx+b\)

  • standard form

    \(Ax+By=C\)

  • point-slope form

    \(y=m\left(x-x_0\right)+y_0\)

We have studied two special types of line:

  • horizontal line: \(y=k\)

  • vertical line: \(x=h\)

We have practiced three ways to graph a line:

  • building a table of \(x\)- and \(y\)-values

  • plotting one point (often the \(y\)-intercept) and drawing slope triangles

  • plotting its \(x\)-intercept and \(y\)-intercept

Subsection4.9.1Graphing Lines in Slope-Intercept Form

In the following examples we will graph \(y=-2x+1\text{,}\) which is in slope-intercept form (4.5.1), with different methods and compare them.

Example4.9.2Building a Table of \(x\)- and \(y\)-values

First, we will graph \(y=-2x+1\) by building a table of values. In theory this method can be used for any type of equation, linear or not. Every student must feel comfortable with building a table of values based on an equation.

\(x\)-value \(y\)-value Point
\(\highlight{-2}\) \(y=-2(\substitute{-2})+1=5\) \((-2,5)\)
\(\highlight{-1}\) \(y=-2(\substitute{-1})+1=3\) \((-1,3)\)
\(\highlight{0}\) \(y=-2(\substitute{0})+1=1\) \((0,1)\)
\(\highlight{1}\) \(y=-2(\substitute{1})+1=-1\) \((1,-1)\)
\(\highlight{2}\) \(y=-2(\substitute{2})+1=-3\) \((2,-3)\)
This is a graph of the line y=-2x+1. The following points on the line are plotted: (-2,5),(-1,3),(0,1),(1,-1),(2,-3).
Table4.9.3Table for \(y=-2x+1\)
Figure4.9.4Graphing \(y=-2x+1\) by Building a Table of Values
Example4.9.5Using Slope Triangles

Although making a table is straightforward, the slope triangle method is both faster and reinforces the true meaning of slope. In the slope triangle method, we first identify some point on the line. With a line in slope-intercept form (4.5.1), we know the \(y\)-intercept, which is \((0,1)\text{.}\) Then, we can draw slope triangles in both directions to find more points.

This is a graph of the line y=-2x+1. The following points on the line are plotted: (-2,5),(-1,3),( 0,1),(1,-1),(2,-3). There are a few slope triangles. One starts at (-2,5), passes (-1,5), and ends at (-1,3). One    starts at (-1,3), passes (0,3), and ends at (0,1). One starts at (0,1), passes (1,1), and ends at (1,-1). One starts   at (1,-1), passes (2,-1), and ends at (2,-3).
Figure4.9.6Marking a point and some slope triangles
Figure4.9.7Graphing \(y=-2x+1\) by slope triangles

Compared to the table method, the slope triangle method:

  • is less straightforward
  • doesn't take the time and space to make a table
  • doesn't involve lots of calculations where you might make a human error
  • shows slope triangles, which reinforces the meaning of slope
Example4.9.8Using intercepts

If we use the \(x\)- and \(y\)-intercepts to plot \(y=-2x+1\text{,}\) we have some calculation to do. While it is apparent that the \(y\)-intercept is at \((0,1)\text{,}\) where is the \(x\)-intercept? Here are two methods to find it.

Set \(y=0\text{.}\)
Factor out the coefficient of \(x\text{.}\)
\begin{align*} y\amp=-2x+1\\ \substitute{0}\amp=-2x+1\\ 0\subtractright{1}\amp=-2x\\ -1\amp=-2x\\ \divideunder{-1}{-2}\amp=x\\ \frac{1}{2}\amp=x \end{align*}
\begin{align*} y\amp=-2x+1\\ y\amp=-2x+(\highlight{-2})\left(\highlight{-\frac{1}{2}}\right)1\\ y\amp=-2\left(x+\left(-\frac{1}{2}\right)1\right)\\ y\amp=-2\left(\highlight{x}-\frac{1}{2}\right) \end{align*}

And now it is easy to see that substituting \(x=\frac{1}{2}\) would make \(y=0\text{.}\)

So the \(x\)-intercept is at \(\left(\frac{1}{2},0\right)\text{.}\) Plotting both intercepts:

Figure4.9.9Marking intercepts
Figure4.9.10Graphing \(y=-2x+1\) by slope triangles

This worked, but here are some observations about why this method is not the greatest.

  • We had to plot a point with fractional coordinates.
  • We only plotted two points and they turned out very close to each other, so even the slightest inaccuracy in our drawing skills could result in a line that is way off.

When a line is presented in slope-intercept form (4.5.1), our opinion is that the slope triangle method is the best choice for making its graph.

Subsection4.9.2Graphing Lines in Standard Form

In the following examples we will graph \(3x+4y=12\text{,}\) which is in standard form (4.7.1), with different methods and compare them.

Example4.9.11Building a Table of \(x\)- and \(y\)-values

To make a table, we could substitute \(x\) for various numbers and use algebra to find the corresponding \(y\)-values. Let's start with \(x=-2\text{,}\) planning to move on to \(x=-1,0,1,2\text{.}\)

\begin{align*} 3x+4y\amp=12\\ 3(\substitute{-2})+4y\amp=12\\ -6+4y\amp=12\\ 4y\amp=12\addright{6}\\ 4y\amp=18\\ y\amp=\divideunder{18}{4}\\ y\amp=\frac{9}{2} \end{align*}

The first point we found is \(\left(-2,\frac{9}{2}\right)\text{.}\) This has been a lot of calculation, and we ended up with a fraction we will have to plot. And we have to repeat this process a few more times to get more points for the table. The table method is generally not a preferred way to graph a line in standard form (4.7.1). Let's look at other options.

Example4.9.12Using intercepts

Next, we will try graphing \(3x+4y=12\) using intercepts. We set up a small table to record the two intercepts:

\(x\)-value \(y\)-value Intercept
\(x\)-intercept \(0\)
\(y\)-intercept \(0\)

We have to calculate the line's \(x\)-intercept by substituting \(y=0\) into the equation:

\begin{align*} 3x+4y\amp=12\\ 3x+4(\substitute{0})\amp=12\\ 3x\amp=12\\ x\amp=\divideunder{12}{3}\\ x\amp=4 \end{align*}

And similarly for the \(y\)-intercept:

\begin{align*} 3x+4y\amp=12\\ 3(\substitute{0})+4y\amp=12\\ 4y\amp=12\\ y\amp=\divideunder{12}{4}\\ y\amp=3 \end{align*}

So the line's \(x\)-intercept is at \((4,0)\) and its \(y\)-intercept is at \((0,3)\text{.}\) Now we can complete the table and then graph the line:

\(x\)-value \(y\)-value Intercepts
\(x\)-intercept \(4\) \(0\) \((4,0)\)
\(y\)-intercept \(0\) \(3\) \((0,3)\)
This is a grid with the graph of line 3x+4y=12. The following points are ploted: (0,3), (4,0).
Table4.9.13Intercepts of \(3x+4y=12\)
Figure4.9.14Graph of \(3x+4y=12\)
Example4.9.15

We can always rearrange \(3x+4y=12\) into slope-intercept form (4.5.1), and then graph it with the slope triangle method:

\begin{align*} 3x+4y\amp=12\\ 4y\amp=12\subtractright{3x}\\ 4y\amp=-3x+12\\ y\amp=\divideunder{-3x+12}{4}\\ y\amp=-\frac{3}{4}x+3 \end{align*}

With the \(y\)-intercept at \((0,3)\) and slope \(-\frac{3}{4}\text{,}\) we can graph the line using slope triangles:

This is a graph of the line 3x+4y=12. The following points on the line are plotted: (-4,6),(0,3),(4,0). There  are  two slope triangles. One starts at (-4,6), passes (0,6), and ends at (0,3). One starts at (0,3), passes (4,3), and  ends at   (4,0).
Figure4.9.16Graphing \(3x+4y=12\) using slope triangles

Compared with the intercepts method, the slope triangle method takes more time, but shows more points with slope triangles, and thus a more accurate graph. Also sometimes (as with Example 4.7.16) when we graph a standard form equation like \(2x-3y=0\text{,}\) the intercepts method doesn't work because both intercepts are actually at the same point, and we have to resort to something else like slope triangles anyway.

Here are some observations about graphing a line equation that is in standard form (4.7.1):

  • The intercepts method might be the quickest approach.
  • The intercepts method only tells us two intercepts of the line. When we need to know more information, like the line's slope, and get a more accurate graph, we should spend more time and use the slope triangle method.
  • When \(C=0\) in a standard form equation (4.7.1) we have to use something else like slope triangles anyway.

Subsection4.9.3Graphing Lines in Point-Slope Form

When we graph a line in point-slope form (4.6.1) like \(y=\frac{2}{3}(x+1)+3\text{,}\) the slope triangle method is the obvious choice. We can see a point on the line, \((-1,3)\text{,}\) and the slope is apparent: \(\frac{2}{3}\text{.}\) Here is the graph:

This is a graph of the line y-3=2/3(x+1). The following points on the line are plotted: (-4,1),(-1,3),(2,5). There are two slope triangles. One starts at (-4,1), passes (-1,1), and ends at (-1,3). One starts at (-1,3), passes (2,3), and ends at (2,5).
Figure4.9.17Graphing \(y-3=\frac{2}{3}(x+1)\) using slope triangles

Other graphing methods would take more work and miss the purpose of point-slope form (4.6.1). To graph a line in point-slope form (4.6.1), we recommend always using slope triangles.

Subsection4.9.4Graphing Horizontal and Vertical Lines

We learned in Section 4.8 that equations in the form \(x=h\) and \(y=k\) make vertical and horizontal lines. But perhaps you will one day find yourself not remembering which is which. Making a table and plotting points can quickly remind you which type of equation makes which type of line. Let's build a table for \(y=2\) and another one for \(x=-3\text{:}\)

\(x\)-value \(y\)-value Point
\(0\) \(2\) \((0,2)\)
\(1\) \(2\) \((1,2)\)
\(x\)-value \(y\)-value Point
\(-3\) \(0\) \((-3,0)\)
\(-3\) \(1\) \((-3,1)\)
Table4.9.18Table of Data for \(y=2\)
Table4.9.19Table of Data for \(x=-3\)

With two points on each line, we can graph them:

This is a graph of y=2 and x=-3. For y=2, the points (0,2) and (1,2) are plotted. For x=-3, the points (-3,0) and (-3,1) are plotted.
Figure4.9.20Graphing \(y=2\) and \(x=-3\)

Subsection4.9.5Exercises

Graphing by Table

1

Use a table to make a plot of \(y=4x+3\text{.}\)

2

Use a table to make a plot of \(y=-5x-1\text{.}\)

3

Use a table to make a plot of \(y=-\frac{3}{4}x-1\text{.}\)

4

Use a table to make a plot of \(y=\frac{5}{3}x+3\text{.}\)

Graphing Standard Form Equations

5

First find the \(x\)- and \(y\)-intercepts of the line with equation \(6x+5y=-90\text{.}\) Then use your results to graph the line.

6

First find the \(x\)- and \(y\)-intercepts of the line with equation \(2x-3y=-6\text{.}\) Then use your results to graph the line.

7

First find the \(x\)- and \(y\)-intercepts of the line with equation \(3x+y=-9\text{.}\) Then use your results to graph the line.

8

First find the \(x\)- and \(y\)-intercepts of the line with equation \(-15x+3y=-3\text{.}\) Then use your results to graph the line.

9

First find the \(x\)- and \(y\)-intercepts of the line with equation \(4x+3y=-3\text{.}\) Then use your results to graph the line.

10

First find the \(x\)- and \(y\)-intercepts of the line with equation \(-4x-5y=5\text{.}\) Then use your results to graph the line.

11

First find the \(x\)- and \(y\)-intercepts of the line with equation \(5x-3y=0\text{.}\) Then use your results to graph the line.

12

First find the \(x\)- and \(y\)-intercepts of the line with equation \(2x+9y=0\text{.}\) Then use your results to graph the line.

Graphing Slope-Intercept Equations

13

Use the slope and \(y\)-intercept from the line \(y=-5x\) to plot the line. Use slope triangles.

14

Use the slope and \(y\)-intercept from the line \(y=3x-6\) to plot the line. Use slope triangles.

15

Use the slope and \(y\)-intercept from the line \(y=-\frac{2}{5}x+2\) to plot the line. Use slope triangles.

16

Use the slope and \(y\)-intercept from the line \(y=\frac{10}{3}x-3\) to plot the line. Use slope triangles.

Graphing Horizontal and Vertical Lines

17

Plot the line \(y=1\text{.}\)

18

Plot the line \(x=-8\text{.}\)

Choosing the Best Method to Graph Lines

19

Use whatever method you think best to plot \(y=2x+2\text{.}\)

20

Use whatever method you think best to plot \(y=-\frac{3}{4}x-1\text{.}\)

21

Use whatever method you think best to plot \(y=-\frac{3}{4}(x-5)+2\text{.}\)

22

Use whatever method you think best to plot \(3x+2y=6\text{.}\)

23

Use whatever method you think best to plot \(3x-4y=0\text{.}\)

24

Use whatever method you think best to plot \(x=-3\text{.}\)