# 11.3: Graphing with Intercepts (Part 2)

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

### Choose the Most Convenient Method to Graph a Line

While we could graph any linear equation by plotting points, it may not always be the most convenient method. This table shows six of equations we’ve graphed in this chapter, and the methods we used to graph them.

Equation | Method | |
---|---|---|

#1 | y = 2x + 1 | Plotting points |

#2 | y = \(\frac{1}{2}\)x + 3 | Plotting points |

#3 | x = −7 | Vertical line |

#4 | y = 4 | Horizontal line |

#5 | 2x + y = 6 | Intercepts |

#6 | 4x − 3y = 12 | Intercepts |

What is it about the form of equation that can help us choose the most convenient method to graph its line?

Notice that in equations #1 and #2, y is isolated on one side of the equation, and its coefficient is 1. We found points by substituting values for x on the right side of the equation and then simplifying to get the corresponding y- values.

Equations #3 and #4 each have just one variable. Remember, in this kind of equation the value of that one variable is constant; it does not depend on the value of the other variable. Equations of this form have graphs that are vertical or horizontal lines.

In equations #5 and #6, both x and y are on the same side of the equation. These two equations are of the form Ax + By = C . We substituted y = 0 and x = 0 to find the x- and y- intercepts, and then found a third point by choosing a value for x or y.

This leads to the following strategy for choosing the most convenient method to graph a line.

HOW TO: CHOOSE THE MOST CONVENIENT METHOD TO GRAPH A LINE

Step 1. If the equation has only one variable. It is a vertical or horizontal line.

- x = a is a vertical line passing through the x-axis at a.
- y = b is a horizontal line passing through the y-axis at b.

Step 2. If y is isolated on one side of the equation. Graph by plotting points.

- Choose any three values for x and then solve for the corresponding y- values.

Step 3. If the equation is of the form Ax + By = C, find the intercepts.

- Find the x- and y- intercepts and then a third point.

Example 11.29:

Identify the most convenient method to graph each line: (a) y = −3 (b) 4x−6y = 12 (c) x = 2 (d) y = 2 5 x−1

##### Solution

(a) y = −3

This equation has only one variable, y. Its graph is a horizontal line crossing the y-axis at −3.

(b) 4x−6y = 12

This equation is of the form Ax + By = C. Find the intercepts and one more point.

(c) x = 2

There is only one variable, x. The graph is a vertical line crossing the x-axis at 2.

(d) y = \(\frac{2}{5}\)x−1

Since y is isolated on the left side of the equation, it will be easiest to graph this line by plotting three points.

Exercise 11.56:

Identify the most convenient method to graph each line: (a) 3x + 2y = 12 (b) y = 4 (c) y = \(\frac{1}{5}\)x−4 (d) x = −7

Exercise 11.57:

Identify the most convenient method to graph each line: (a) x = 6 (b) y = \(− \frac{3}{4}\)x + 1 (c) y = −8 (d) 4x−3y = −1

ACCESS ADDITIONAL ONLINE RESOURCES

### Practice Makes Perfect

#### Identify the Intercepts on a Graph

In the following exercises, find the x- and y- intercepts.

#### Find the x and y Intercepts from an Equation of a Line

In the following exercises, find the intercepts.

- x + y = 4
- x + y = 3
- x + y = −2
- x + y = −5
- x − y = 5
- x − y = 1
- x − y = −3
- x − y = −4
- x + 2y = 8
- x + 2y = 10
- 3x + y = 6
- 3x + y = 9
- x−3y = 12
- x−2y = 8
- 4x − y = 8
- 5x − y = 5
- 2x + 5y = 10
- 2x + 3y = 6
- 3x−2y = 12
- 3x−5y = 30
- y = \(\frac{1}{3}\)x−1
- y = \(\frac{1}{4}\)x−1
- y = \(\frac{1}{5}\)x + 2
- y = \(\frac{1}{3}\) x + 4
- y = 3x
- y = −2x
- y = −4x
- y = 5x

#### Graph a Line Using the Intercepts

In the following exercises, graph using the intercepts.

- −x + 5y = 10
- −x + 4y = 8
- x + 2y = 4
- x + 2y = 6
- x + y = 2
- x + y = 5
- x + y = −3
- x + y = −1
- x − y = 1
- x − y = 2
- x − y = −4
- x − y = −3
- 4x + y = 4
- 3x + y = 3
- 3x − y = −6
- 2x − y = −8
- 2x + 4y = 12
- 3x + 2y = 12
- 3x−2y = 6
- 5x−2y = 10
- 2x−5y = −20
- 3x−4y = −12
- y = −2x
- y = −4x
- y = x
- y = 3x

#### Choose the Most Convenient Method to Graph a Line

In the following exercises, identify the most convenient method to graph each line.

- x = 2
- y = 4
- y = 5
- x = −3
- y = −3x + 4
- y = −5x + 2
- x − y = 5
- x − y = 1
- y = \(\frac{2}{3}\)x−1
- y = \(\frac{4}{5}\)x−3
- y = −3
- y = −1
- 3x−2y = −12
- 2x−5y = −10
- y = \(− \frac{1}{4}\)x + 3
- y = \(− \frac{1}{3}\)x + 5

### Everyday Math

**Road trip**Damien is driving from Chicago to Denver, a distance of 1,000 miles. The x-axis on the graph below shows the time in hours since Damien left Chicago. The y-axis represents the distance he has left to drive.- Find the x- and y- intercepts.
- Explain what the x- and y- intercepts mean for Damien.

**Road trip**Ozzie filled up the gas tank of his truck and went on a road trip. The x-axis on the graph shows the number of miles Ozzie drove since filling up. The y-axis represents the number of gallons of gas in the truck’s gas tank.- Find the x- and y- intercepts.
- Explain what the x- and y- intercepts mean for Ozzie.

### Writing Exercises

- How do you find the x-intercept of the graph of 3x−2y = 6?
- How do you find the y-intercept of the graph of 5x − y = 10?
- Do you prefer to graph the equation 4x + y = −4 by plotting points or intercepts? Why?
- Do you prefer to graph the equation y = \(\frac{2}{3}\)x − 2 by plotting points or intercepts? Why?

### Self Check

(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

(b) What does this checklist tell you about your mastery of this section? What steps will you take to improve?

### Contributors

Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at http://cnx.org/contents/fd53eae1-fa2...49835c3c@5.191."