Learn on PengiSaxon Algebra 2Chapter 2: Lessons 11-20, Investigation 2

Lesson 13: Graphing Linear Equations I

In this Grade 10 Saxon Algebra 2 lesson, students learn to graph linear equations in two variables using three methods: constructing a table of values, plotting x- and y-intercepts, and applying slope-intercept form (y = mx + b). Students also practice calculating slope using the rise-over-run formula and classifying lines as rising, falling, horizontal, or vertical based on slope values.

Section 1

πŸ“˜ Graphing Linear Equations I

New Concept

A linear equation written in the form y=mx+by = mx + b is written in slope-intercept form, where mm is the slope and bb is the yy-intercept.

What’s next

Next, you’ll master graphing these equations using solution tables, intercepts, and the powerful slope-intercept form.

Section 2

Graphing a Linear Equation Using Intercepts

Find the xx-intercept by setting y=0y = 0 and solving for xx. Then, find the yy-intercept by setting x=0x = 0 and solving for yy. Plot these two points and draw a straight line through them.

To graph 4xβˆ’2y=84x - 2y = 8: For the xx-intercept, set y=0y=0 to get 4x=84x=8, so x=2x=2. The point is (2,0)(2, 0). For the yy-intercept, set x=0x=0 to get βˆ’2y=8-2y=8, so y=βˆ’4y=-4. The point is (0,βˆ’4)(0, -4). Plot and connect (2,0)(2, 0) and (0,βˆ’4)(0, -4).
To graph βˆ’x+3y=6-x + 3y = 6: For the xx-intercept, set y=0y=0 to get βˆ’x=6-x=6, so x=βˆ’6x=-6. The point is (βˆ’6,0)(-6, 0). For the yy-intercept, set x=0x=0 to get 3y=63y=6, so y=2y=2. The point is (0,2)(0, 2). Plot and connect (βˆ’6,0)(-6, 0) and (0,2)(0, 2).

Why hunt for a bunch of points when you only need two? The intercepts are the line's special 'touchdown' spots on the axes. Finding them is the fastest way to get your line on the graph and score a perfect plot!

Section 3

Slope of a Line

The slope mm of a line through points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is the ratio of vertical change (rise) to horizontal change (run).

m=y2βˆ’y1x2βˆ’x1=riserunm = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}

For points (βˆ’1,2)(-1, 2) and (1,6)(1, 6): m=6βˆ’21βˆ’(βˆ’1)=42=2m = \frac{6 - 2}{1 - (-1)} = \frac{4}{2} = 2. The positive slope means the line rises.
For points (2,5)(2, 5) and (4,1)(4, 1): m=1βˆ’54βˆ’2=βˆ’42=βˆ’2m = \frac{1 - 5}{4 - 2} = \frac{-4}{2} = -2. The negative slope means the line falls.
For points (βˆ’3,4)(-3, 4) and (5,4)(5, 4): m=4βˆ’45βˆ’(βˆ’3)=08=0m = \frac{4 - 4}{5 - (-3)} = \frac{0}{8} = 0. The zero slope means the line is horizontal.

Slope is the secret code for a line's steepness and direction! It’s like a recipe that tells you how many steps to go up or down for every step you take sideways. A positive slope climbs uphill, while a negative one slides downhill.

Section 4

The Slope-Intercept Form of a Linear Equation

A linear equation written in the form y=mx+by = mx + b is in slope-intercept form. Here, mm represents the slope of the line, and bb represents the yy-intercept.

For y=βˆ’13x+2y = -\frac{1}{3}x + 2, the slope is βˆ’13-\frac{1}{3} and the y-intercept is 22. Start at (0,2)(0, 2), then move down 1 unit and right 3 units to find the next point at (3,1)(3, 1).
To graph 4x+2y=84x + 2y = 8, first solve for yy: 2y=βˆ’4x+82y = -4x + 8, which gives y=βˆ’2x+4y = -2x + 4. The slope is βˆ’2-2 and the y-intercept is 44. Start at (0,4)(0, 4) and move down 2, right 1.

This form is a graphing treasure map! The equation literally hands you your starting point, bb (where the line crosses the y-axis), and your directions, mm (the slope), to find the next point on the map. No tables, no extra math, just pure graphing action!

Book overview

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Chapter 2: Lessons 11-20, Investigation 2

  1. Lesson 1

    Lesson 11: Understanding Polynomials

  2. Lesson 2

    Lesson 12: Solving Inverse Variation Problems

  3. Lesson 3

    Lab 3: Graphing Calculator: Calculating Points on a Graph

  4. Lesson 4Current

    Lesson 13: Graphing Linear Equations I

  5. Lesson 5

    Lesson 14: Finding Determinants

  6. Lesson 6

    Lesson 15: Solving Systems of Equations by Graphing

  7. Lesson 7

    Lesson 16: Using Cramer's Rule

  8. Lesson 8

    LAB 4: Graphing Calculator: Changing the Line and Window of a Graph

  9. Lesson 9

    Lesson 17: Solving Equations and Inequalities with Absolute Value (Exploration: Transforming f(x) = |x|)

  10. Lesson 10

    Lesson 18: Calculating with Units of Measure

  11. Lesson 11

    Lesson 19: Multiplying Polynomials

  12. Lesson 12

    Lesson 20: Performing Operations with Functions

  13. Lesson 13

    Investigation 2: Solving Parametric Equations

Lesson overview

Expand to review the lesson summary and core properties.

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Section 1

πŸ“˜ Graphing Linear Equations I

New Concept

A linear equation written in the form y=mx+by = mx + b is written in slope-intercept form, where mm is the slope and bb is the yy-intercept.

What’s next

Next, you’ll master graphing these equations using solution tables, intercepts, and the powerful slope-intercept form.

Section 2

Graphing a Linear Equation Using Intercepts

Find the xx-intercept by setting y=0y = 0 and solving for xx. Then, find the yy-intercept by setting x=0x = 0 and solving for yy. Plot these two points and draw a straight line through them.

To graph 4xβˆ’2y=84x - 2y = 8: For the xx-intercept, set y=0y=0 to get 4x=84x=8, so x=2x=2. The point is (2,0)(2, 0). For the yy-intercept, set x=0x=0 to get βˆ’2y=8-2y=8, so y=βˆ’4y=-4. The point is (0,βˆ’4)(0, -4). Plot and connect (2,0)(2, 0) and (0,βˆ’4)(0, -4).
To graph βˆ’x+3y=6-x + 3y = 6: For the xx-intercept, set y=0y=0 to get βˆ’x=6-x=6, so x=βˆ’6x=-6. The point is (βˆ’6,0)(-6, 0). For the yy-intercept, set x=0x=0 to get 3y=63y=6, so y=2y=2. The point is (0,2)(0, 2). Plot and connect (βˆ’6,0)(-6, 0) and (0,2)(0, 2).

Why hunt for a bunch of points when you only need two? The intercepts are the line's special 'touchdown' spots on the axes. Finding them is the fastest way to get your line on the graph and score a perfect plot!

Section 3

Slope of a Line

The slope mm of a line through points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is the ratio of vertical change (rise) to horizontal change (run).

m=y2βˆ’y1x2βˆ’x1=riserunm = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}

For points (βˆ’1,2)(-1, 2) and (1,6)(1, 6): m=6βˆ’21βˆ’(βˆ’1)=42=2m = \frac{6 - 2}{1 - (-1)} = \frac{4}{2} = 2. The positive slope means the line rises.
For points (2,5)(2, 5) and (4,1)(4, 1): m=1βˆ’54βˆ’2=βˆ’42=βˆ’2m = \frac{1 - 5}{4 - 2} = \frac{-4}{2} = -2. The negative slope means the line falls.
For points (βˆ’3,4)(-3, 4) and (5,4)(5, 4): m=4βˆ’45βˆ’(βˆ’3)=08=0m = \frac{4 - 4}{5 - (-3)} = \frac{0}{8} = 0. The zero slope means the line is horizontal.

Slope is the secret code for a line's steepness and direction! It’s like a recipe that tells you how many steps to go up or down for every step you take sideways. A positive slope climbs uphill, while a negative one slides downhill.

Section 4

The Slope-Intercept Form of a Linear Equation

A linear equation written in the form y=mx+by = mx + b is in slope-intercept form. Here, mm represents the slope of the line, and bb represents the yy-intercept.

For y=βˆ’13x+2y = -\frac{1}{3}x + 2, the slope is βˆ’13-\frac{1}{3} and the y-intercept is 22. Start at (0,2)(0, 2), then move down 1 unit and right 3 units to find the next point at (3,1)(3, 1).
To graph 4x+2y=84x + 2y = 8, first solve for yy: 2y=βˆ’4x+82y = -4x + 8, which gives y=βˆ’2x+4y = -2x + 4. The slope is βˆ’2-2 and the y-intercept is 44. Start at (0,4)(0, 4) and move down 2, right 1.

This form is a graphing treasure map! The equation literally hands you your starting point, bb (where the line crosses the y-axis), and your directions, mm (the slope), to find the next point on the map. No tables, no extra math, just pure graphing action!

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 2: Lessons 11-20, Investigation 2

  1. Lesson 1

    Lesson 11: Understanding Polynomials

  2. Lesson 2

    Lesson 12: Solving Inverse Variation Problems

  3. Lesson 3

    Lab 3: Graphing Calculator: Calculating Points on a Graph

  4. Lesson 4Current

    Lesson 13: Graphing Linear Equations I

  5. Lesson 5

    Lesson 14: Finding Determinants

  6. Lesson 6

    Lesson 15: Solving Systems of Equations by Graphing

  7. Lesson 7

    Lesson 16: Using Cramer's Rule

  8. Lesson 8

    LAB 4: Graphing Calculator: Changing the Line and Window of a Graph

  9. Lesson 9

    Lesson 17: Solving Equations and Inequalities with Absolute Value (Exploration: Transforming f(x) = |x|)

  10. Lesson 10

    Lesson 18: Calculating with Units of Measure

  11. Lesson 11

    Lesson 19: Multiplying Polynomials

  12. Lesson 12

    Lesson 20: Performing Operations with Functions

  13. Lesson 13

    Investigation 2: Solving Parametric Equations