C ollege A lgebra Linear and Quadratic Functions (Chapter2) 1.

College Algebra

Linear and Quadratic Functions(Chapter2)

1

2

Objectives

Chapter2

Cover the topics in ( Section 2.1:Linear functions)

After completing these sections, you should be able to:

1. Calculate and Interpret the Slope of a Line2. Graph Lines Given a Point and the Slope3. Use the Point-Slope Form of a Line4. Find the Equation of a Line Given Two Points5. Write the Equation of a Line in Slope-Intercept From and in General Form.

3

Slope of a Line

Chapter2

Let and be two distinct points with . The slope m of the non-vertical line L containing P and Q is defined by the formula

11, yxP 22 , yxQ

21 xx

my yx x

x x

2 1

2 11 2

If , L is a vertical line and the slope m of L is undefined (since this results in division by 0).

21 xx

4

Slope of a Line

Chapter2

x x2 1

y y2 1P = ( , )x y1 1

Q = ( , )x y2 2

y

x

Slope can be though of as the ratio of the vertical change ( ) to the horizontal change ( (, often termed “rise over run”.

y y2 1 x x2 1

5

Slope of a Line

Chapter2

Definition of Slope

The slope of the line through the distinct points (x1, y1) and (x2, y2) is

where x2 – x1 = 0.

Definition of Slope

The slope of the line through the distinct points (x1, y1) and (x2, y2) is

where x2 – x1 = 0.

Change in y

Change in x=

Rise

Run=y2 – y1

x2 – x1

(x1, y1)

x1

y1

x2

y2

(x2, y2)Risey2 – y1

Runx2 – x1

x

y

Definition of Slope

6

Slope of a Line

Chapter2

The Possibilities for a Line’s SlopePositive Slope

x

y

m > 0

Line rises from left to right.

Zero Slope

x

y

m = 0

Line is horizontal.m is

undefined

Undefined Slope

x

y

Line is vertical.

Negative Slope

x

y

m < 0

Line falls from left to right.

7

Slope of a Line

Chapter2

P = ( , )x y1 1

Q = ( , )x y1 2

Ly

x

If , then is zero and the slope is undefined. Plotting the two points results in the graph of a vertical line with the equation .

21 xx x x2 1

1xx

8

Slope of a Line

Chapter2

Example(1): Find the slope of the line joining the points (3,8) and (-1,2).

x y x y1 1 2 23 8 1 2, , , ,

my yx x

2 1

2 1

m 2 81 3

64

32

9

Slope of a Line

Chapter2

Example(2): Draw the graph of the line passing through (1,4) with a slope of -3/2.

Step 1: Plot the given point.

Step 2: Use the slope to find another point on the line (vertical change = -3, horizontal change = 2).

y

x

(1,4)

2

-3

(3,1)

10

Slope of a Line

Chapter2

Example(3): Draw the graph of the equation x = 2.

y

x

x = 2

11

Equation of a Line

Chapter2

When writing an equation of a line, keep in mind that you ALWAYS need two pieces of information when you go to write an equation:

1. ANY point on the line 2. Slope

Equation of a Line

Chapter2

Point-Slope Form of an Equation of a Line

An equation of a non-vertical line of slope m that passes through the point (x1, y1) is:

y y m x x 1 1

13

Equation of a Line

Chapter2

Example(4): Find an equation of a line with slope -2 passing through (-1,5).

m x y 2 1 51 1 and , ,

y y m x x 1 1

y x 5 2 1

y x 5 2 2

y x 2 3

14

Equation of a Line

Chapter2

15

Equation of a Line

Chapter2

Slope/Intercept Equation of a Line

In this form, m represents the slope and b represents the y-intercept of the line.

16

Equation of a Line

Chapter2

17

Equation of a Line

Chapter2

Equation of a Horizontal Line

A horizontal line is given by an equation of the form y = b where b is the y-intercept.

18

Equation of a Line

Chapter2

Equation of a Vertical Line

A vertical line is given by an equation of the form x = awhere a is the x-intercept.

19

Equation of a Line

Chapter2

20

Equation of a Line

Chapter2

21

Equation of a Line

Chapter2

22

Equation of a Line

Chapter2

23

Equation of a Line

Chapter2

The equation of a line L is in general form with it is written as

Ax By C 0

where A, B, and C are three real numbers and A and B are not both 0.

The equation of a line L is in slope-intercept form with it is written as

y = mx + b

where m is the slope of the line and (0,b) is the y-intercept.

24

Equation of a Line

Chapter2

Example: Find the slope m and y-intercept (0,b) of the graph of the line 3x - 2y + 6 = 0.

3x - 2y + 6 = 0

-2y = -3x - 6

y x 32

3

m 32

b 3

25

Equation of a Line

Chapter2

Example: Write the point-slope form of the equation of the line passing through (-1,3) with a slope of 4. Then solve the equation for y.

Solution We use the point-slope equation of a line with m = 4, x1= -1, and

y1 = 3.

This is the point-slope form of the equation.y – y1 = m(x – x1)

Substitute the given values.y – 3 = 4[x – (-1)]

We now have the point-slope form of the equation for the given line.

y – 3 = 4(x + 1)

We can solve the equation for y by applying the distributive property.

y – 3 = 4x + 4

y = 4x + 7 Add 3 to both sides.

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Equation of a Line

Chapter2

Example: Graph the line whose equation is y = 2/3 x + 2.

Solution The equation of the line is in the form y = mx + b. We can find the slope, m, by identifying the coefficient of x. We can find the y-intercept, b, by identifying the constant term.

y = 2/3 x + 2 23

The slope is 2/3.

The y-intercept is 2.

We plot the second point on the line by starting at (0, 2), the first point. Then move 2 units up (the rise) and 3 units to the right (the run). This gives us a second point at (3, 4).

-5 -4 -3 -2 -1 1 2 3 4 5

5

4

3

2

1

-1

-2

-3

-4

-5

We need two points in order to graph the line. We can use the y-intercept, 2, to obtain the first point (0, 2). Plot this point on the y-axis.

27

Equation of a Line

Chapter2

Equation of a Horizontal Line

A horizontal line is given by an equation of the formy = bwhere b is the y-intercept.

Equation of a Vertical Line

A vertical line is given by an equation of the formx = awhere a is the x-intercept.

28

Equation of a Line

Chapter2

29

Equation of a Line

Chapter2

Equations of Lines

Point-slope form: y – y1 = m(x – x1) Slope-intercept form: y = m x + b

Horizontal line: y = b Vertical line: x = a

General form: Ax + By + C = 0

30

Parallel and Perpendicular Lines

Chapter2

Definitions: Parallel Lines

Two lines are said to be parallel if they do not have any points in common.

Two distinct non-vertical lines are parallel if and only if they have the same slope and have different y-intercepts.

31

Parallel and Perpendicular Lines

Chapter2

32

Parallel and Perpendicular Lines

Chapter2

Definitions: Perpendicular Lines

Two lines are said to be perpendicular if they intersect at a right angle.

Two non-vertical lines are perpendicular if and only if the product of their slopes is -1.

121 mm1

2

1

mm

33

Parallel and Perpendicular Lines

Chapter2

34

Parallel and Perpendicular Lines

Chapter2

Example: Find the equation of the line parallel to y = -3x + 5 passing through (1,5).

Since parallel lines have the same slope, the slope of the parallel line is m = -3.

y y m x x 1 1

y x 5 3 1

y x 5 3 3

y x 3 8

35

Parallel and Perpendicular Lines

Chapter2

Example: Find the equation of the line perpendicular to y = -3x + 5 passing through (1,5).

Slope of perpendicular line:

13

13

y y m x x 1 1

y x 513

1

y x 513

13

y x 13

143

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Parallel and Perpendicular Lines

Chapter2

Slope and Parallel Lines

• If two non-vertical lines are parallel, then they have the same slope.

• If two distinct non-vertical lines have the same slope, then they are parallel.

• Two distinct vertical lines, both with undefined slopes, are parallel.

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Parallel and Perpendicular Lines

Chapter2

Example: Write an equation of the line passing through (-3, 2) and parallel to the line whose equation is y = 2x + 1. Express the equation in point-slope form and y-intercept form.

Solution We are looking for the equation of the line shown on the left. Notice that the line passes through the point (-3, 2). Using the point-slope form of the line’s equation, we have x1 = -3 and y1 = 2.

y = 2x + 1

-5 -4 -3 -2 -1 1 2 3 4 5

5

4

3

2

1

-1-2

-3

-4-5

(-3, 2)

Rise = 2

Run = 1y – y1 = m(x – x1)

y1 = 2 x1 = -3

38

Parallel and Perpendicular Lines

Chapter2

Solution Parallel lines have the same slope. Because the slope of the given line is 2, m = 2 for the new equation.

-5 -4 -3 -2 -1 1 2 3 4 5

5

4

3

2

1

-1

-2

-3

-4

-5

(-3, 2)

Rise = 2

Run = 1

y = 2x + 1

y – y1 = m(x – x1)

y1 = 2 m = 2 x1 = -3

Example cont.

39

Parallel and Perpendicular Lines

Chapter2

Solution The point-slope form of the line’s equation is

y – 2 = 2[x – (-3)]

y – 2 = 2(x + 3)

Solving for y, we obtain the slope-intercept form of the equation.

y – 2 = 2x + 6 Apply the distributive property.

y = 2x + 8 Add 2 to both sides. This is the slope-intercept form of the equation.

Example cont.

40

Parallel and Perpendicular Lines

Chapter2

Slope and Perpendicular Lines

Slope and Perpendicular Lines• If two non-vertical lines are perpendicular, then the product of their

slopes is –1.• If the product of the slopes of two lines is –1, then the lines are

perpendicular.• A horizontal line having zero slope is perpendicular to a vertical line

having undefined slope.

Slope and Perpendicular Lines• If two non-vertical lines are perpendicular, then the product of their

slopes is –1.• If the product of the slopes of two lines is –1, then the lines are

perpendicular.• A horizontal line having zero slope is perpendicular to a vertical line

having undefined slope.

Two lines that intersect at a right angle (90°) are said to be perpendicular. There is a relationship between the slopes of perpendicular lines.

90°

41

Parallel and Perpendicular Lines

Chapter2

Example: Find the slope of any line that is perpendicular to the line whose equation is 2x + 4y – 4 = 0.

Solution We begin by writing the equation of the given line in slope-intercept form. Solve for y.

2x + 4y – 4 = 0 This is the given equation.

4y = -2x + 4 To isolate the y-term, subtract x and add 4 on both sides.

Slope is –1/2.

y = -2/4x + 1 Divide both sides by 4.

The given line has slope –1/2. Any line perpendicular to this line has a slope that is the negative reciprocal, 2.