Algebra 2

Chapter 2 Notes

Linear Equations and Functions

1

A RELATION is a mapping or pairing of input values with output values. The set of input values in the DOMAIN and the set of output values is the RANGE. A relation is a FUNCTION if there is exactly one output for each input. It is not a function if at lest one input has more than one output.

( x , y ) = (domain , range ) = ( input , output) = (independent , dependent)

Not a function Yes a function

- 3 3 - 3 3

1 –2 1 1

1 2

4 4 4 –2

A relation is a function if and only if no vertical line intersects the graph of the relationship at more than one point.

NOT a RELATION YES a RELATION 2

Functions and their Graphs 2.1

Y axis

Quadrant I

( + , + )

Quadrant II

( – , + )

Quadrant III

( – , – )

Quadrant IV

( + , – )

X axis

( 0 , 0 )

Ordered pairs in form of ( x , y )

Coordinate Plane

x coordinate is firsty coordinate is second 3

2.1

Graphing equations in 2 variables

1. Construct a table of values

2. Graph enough solutions to recognize a pattern

3. Connect the points with a line or a curve

Graph the function: y = x + 1

y = m x + b LINEAR FUNCTION

f (x) = m x + b FUNCTION NOTATION

x y = x + 1 y

- 2 y = -2 + 1 - 1

- 1 y = -1 + 1 0

0 y = 0 + 1 1

1 y = 1 + 1 2

2 y = 2 + 1 3

y

x

4

2.1Functions and their Graphs

Are these functions linear?. Evaluate when x = – 2

a) f ( x ) = – x2 – 3 x + 5 Not a function because x is to the 2nd power

f ( – 2 ) = – ( – 2 ) 2 – 3 ( – 2 ) + 5

f ( – 2 ) = 7

b) g ( x ) = 2 x + 6 Yes a function because x is to the 1st power

g ( – 2 ) = 2 ( – 2 ) + 6

g (– 2 ) = 2

5

Functions and their Graphs 2.1

SLOPE of a non-vertical line is the ratio of a vertical change (RISE) to a horizontal change (RUN).

Slope of a line:m = y2 – y1 = RISE

x2 – x1 = RUN

(differences in y values)(differences in x values)

•

y2 – y1

RISE

x2 – x1

RUNx

y

( x1 , y1 )

( x2 , y2 )•

6

Slope and Rate of Change 2.1

CLASSIFICATION OF LINES BY SLOPE

A line with a + slope rises from left to right [ m > 0 ]

A line with a – slope falls from left to right [ m < 0 ]

A line with a slope of 0 is horizontal [ m = 0 ]

A line with an undefined slope is vertical [ m = undefined, no slope ]

Positive Slope

Negative Slope

0 Slope

No Slope 7

Slope and Rate of Change 2.2

SLOPES OF PARALLEL & PERPENDICULAR LINES

PARALLEL LINES: the lines are parallel if and only if they have the SAME SLOPE.

m1 = m2

PERPENDICULAR LINES: the lines are perpendicular if and only if their SLOPES are NEGATIVE RECIPROCALS.

m1 = –1 m2

m1 m2 = – 1

or

8

Slope and Rate of Change 2.2

Ex 1:

Find the slope of a line passing through ( – 3, 5 ) and ( 2, 1 )

Let ( x1, y1 ) = ( – 3, 5 ) and ( x2, y2 ) = ( 2, 1 )

y

x

( 2, 1 )

( – 3, 5 )

•

•

5

– 4

Slope of a line:m = y2 – y1 = RISE (differences in y values)

x2 – x1 = RUN (differences in x values)

m = 5 – 1 = 4 – 3 – 2 – 5

m = 1 – 5 = – 4 2 + 3 5

OR

9

Slope and Rate of Change 2.2

Example 2: Without graphing tells if slope rises, falls, horizontal or vertical:

a) ( 3, – 4 ) and ( 1, – 6 ) : m = – 6 – ( – 4 ) = - 2 = 1 m > 0, rises

1 – 3 - 2

b) ( 2, 2 ) and ( – 1, 5 ) : m = 5 – ( – 1 ) = 6 = undefined m = no slope

2 – 2 0

For 2 lines with + slopes, the line with > slope is steeper

For 2 lines with – slopes, the line with slope of > absolute value is steeper:

y

x

m = 1

m = 1/2

m = 3m = - 3m = - 1

m = - 1/2

10

Slope and Rate of Change 2.2

Classifying Perpendicular Lines

Line 1 through ( – 3 , 3 ) and ( 3 , – 1 )

m 1 = – 1 – 3 = – 4 = – 2

3 – ( - 3) 6 3

Line 2 through ( - 2 , - 3 ) and ( 2 , 3 )

m 2 = 3 – ( – 3 ) = 6 = 3

2 – ( – 2 ) 4 2Because m1 m∙ 2 = – 2 3 ∙ = – 1 3 2are negative reciprocals of each other, the lines are perpendicular.

• •

•

•

( –3 , 3 )

( 2, 3 )

( 3 , – 1 )

( –2 , – 3 )

L1

L2

x

y

11

Classifying Lines Using Slopes 2.2

Classifying Parallel Lines

Line 1 through ( – 3 , 3 ) and ( 3 , – 1 )

m 1 = 4 – 1 = 3 = 1

3 – ( - 3) 6 2

Line 2 through ( - 2 , - 3 ) and ( 2 , 3 )

m 2 = 1 – ( – 3 ) = 4 = 1

4 – ( – 4 ) 8 2Because m1 = m2 and the lines are different,then the lines are parallel.

•

•

•

•( -3 , 1 )

( 3, 4 )

( 4 , 1 )

( -4 , - 3 )

L1

L2

x

y

12

Classifying Lines Using Slopes 2.2

Slope Intercept Form of a linear equation is y = m x + b, where m is slope and b is y-intercept

Graphing Equations in slope-intercept form:1. Write the equation in slope-intercept form by solving for y2. Find y-intercept, then plot the point where line crosses the y-axis3. Find the slope and use it to plot a second point on the line.4. Draw a line through the 2 points.

Example 1: Graphing with the slope-intercept

Graph: y = 3 x − 2 4

1. Already in slope-intercept form2. y -intercept is – 2, plot point ( 0 , – 2 )

where the line crosses the y-axis3. Slope is ¾ , so plot 4 units to right,

3 units up, point is ( 4 , 1 )4. Draw a line through the 2 points •

• x

y

3

4( 0 , – 2 )

( 4 , 1 )

13

Quick Graphs of Linear Equations 2.3

Standard Form of a linear equation is ax + by = c

x-intercept of a line is the x-coordinate of the point where the line intersects the x-axis

Graphing Equations in standard form1. Write the equation in standard form2. Find x-intercept, by letting y = 0, solve for x, plot the point where x crosses the x-axis3. Find y-intercept, by letting x = 0, solve for y, plot the point where y crosses the y-axis4. Draw a line through the 2 points.

Method 1 by using Standard Form: Graph 2 x + 3 y = 12

1. Already using standard form: 2 x2. Let y = 0 2 x + 3(0) = 12

2 x = 12 x = 6

x-intercept is at ( 6 , 0 )

3. Let x = 0 2 (0) + 3 y = 12 3 y = 12 y = 4

y-intercept is at ( 0 , 4 )

( 0 , 4 )

( 6 , 0 )

•

•

2 x + 3 y = 12

14

Quick Graphs of Linear Equations 2.3

Method 2 by using Slope-Intercept Form: Graph 2 x + 3 y = 12

1. Change from standard form to slope-intercept form: 2 x + 3 y = 12

-2 x - 2 x 3 y = – 2 x + 12 3 3

y = – 2 x + 4 3

2. Identify , plot y-intercept3. Use slope to plot other point4. Draw a line through the points

( 0 , 4 )

( 3 , 2 )

•

•

3

2

2 x + 3 y = 12y-intercept

slope

Horizontal lines --- graph of y = c is a horizontal line through, (0 , c)

Vertical lines --- graph of x = c is a vertical line through (c, 0)

Graph y = 3 and x = -2

y = 3

x = 3

•

•

( 0 , 3 )

( - 2 , 0 ) x

y

15

Quick Graphs of Linear Equations 2.3

Slope-Intercept Form: y = m x + b

Point-Intercept Form: y – y1 = m ( x – x1 )

Two Points: m = y2 – y1

x2 – x1

* Every non-vertical line has only one slope and one y-intercept

m = slopeb = y-intercept

• Write an equation of line shown• From graph you can see the slope, m = 3

2 • From graph you can see that it intersects at ( 0 ,

−1 ) , so the y-intercepts is b = −1 x

y

( 0 , −1 )

( 2 , 2 )•

• 2

3Equation of the Line is:

y = 3 x − 1 2

16

Writing Equations of Lines 2.4

m ( x1 , y1 ) Writing equations of the line given slope of − 1 and a point, ( 2 , 3 )

2y – y1 = m ( x – x1 )

y – 3 = – 1 ( x – 2 ) 2

y – 3 = – 1 x + 1 2

+ 3 + 3

y = – 1 x + 4 2

Equation of the Line

17

Writing Equations of Lines 2.4

Example 3a: Equation of a line that passes through ( 3 , 2 ) that is perpendicular to the line: y = – 3 x + 2

Perpendicular means m2 = – 1 m1

y – y1 = m2 ( x – x1)

y – 2 = 1 ( x – 3) 3y – 2 = 1 x – 1 3 + 2 + 2y = 1 x + 1 3

Example 3b: Equation of a line that passes through ( 3 , 2 ) that is parallel to the line: y = – 3 x + 2

Parallel means m2 = m1 = – 3 and ( x1 , y1 ) = ( 3 , 2 )

y – y1 = m2 ( x – x1)

y – 2 = – 3 ( x – 3) y – 2 = – 3 x + 9 + 2 + 2y = – 3 x + 11

18

Writing Equations of Perpendicular and Parallel Lines

2.4

Writing Equations given 2 points where ( x1 , y1 ) = ( – 2 , – 1 )

( x2 , y2 ) = ( 3 , 4 )

m = y2 – y1 = 4 – ( – 1 ) = 5 = 1

x2 – x1 3 – ( – 2 ) 5

y – y1 = m ( x – x1)

y – ( – 1 ) = 1 [ x – ( – 2 ) ]

y + 1 = 1 ( x + 2 )

y = x + 1

19

Writing Equations of Lines 2.4

Writing the Equation of the Line

The form you use depends on the information you have

INFO PROVIDED Name of FORM to be used FORM

Slope = 2y - intercept = – 7 Slope – Intercept Form y = m x + b

y = 2 x – 7

Point ( 2, 3 ) m = 4

Point – Slope FormAlso known as

Point – Intercept Form

( x – x 1 ) m = ( y – y 1 )( x – 2 ) 4 = ( y – 3)

Point ( 2, 3 ) Point (– 1, 4 )

Use m = y – y 1

x – x 1 Then use:

Point – Slope Form

m = 4 – 3 = 1– 1 – (– 2 )

( x – 2 ) 1 = ( y – 3)20

2.4

Direct Variation is shown by y = k x , where k ≠ 0 and k is a constant, (that is a single value)

Example 1: Variables x and y vary directly, y = 12 and x = 4

• write and graph an equation for y and x

• find y when x = 5

y = k x

(12 ) = k (4)

3 = k

y = 3 x

y

x

x y 0 0 4 12 5 15 •

••

Example 2x = 6 and y = 3

y = k x3 = k 61 = k2

y = 1 x 2

Example 3x = 9 and y = 15

y = k x15 = k 9

5 = k 3

y = 5 x 3

21

Writing Direct Variation Equations 2.5

A scatter plot is a graph used to determine, whether there is a relationship between paired data.

Positive Correlation Negative Correlation

No Correlation

•

•

••• ••••

•

••• •

••

•

••

••

•

••

•

• • ••

•

•

•••

•

••• •

•

22

Scatter Plots 2.5

A scatter plot is a graph used to determine, whether there is a relationship between paired data

Approximating a best-fitting line: Graphical approach

Step 1: draw a scatter plot of dataStep 2: sketch the line based on the patternStep 3: choose 2 points on the lineStep 4: find the equation of the line that passes through the 2 points

Example 2: Fitting a line to data

Line through two points:

m = 1 − .6 = .4 = .25 2.5 - .9 1.6

Point-slope form:y – y1 = m ( x – x1)y – .6 = .25 ( x – .91)y – .6 = .25 x - .225 + .6 + .6 y = .25 x + .225

0 1 2 3

1.2

.8

.4

.2

••

••

•

• •

••

•

••

••

23

Scatter Plots 2.5

Linear Inequalities in Two Variables can be written in one of the following forms:

ax + by < c ax + by ≤ c ax + by > c ax + by ≥ c

An ordered pair ( x , y ) is a solution of a linear inequality if the inequality is true for values of x , y. Example, ( − 6 , 2 ) is a solution of y ≥ 3 x − 9 because 2 ≥ 3 (− 6) − 9 is true as 2 ≥ − 27

Example 1: Checking Solutions of Inequalities

Ordered pairs Substitute Conclusion

( 0 , 1 ) 2 ( 0 ) + 3 ( 1 ) ≥ 5

3 ≥ 5 False, not a solution

( 4 , − 1 ) 2 (4 ) + 3 (− 1 ) ≥ 5

5 ≥ 5 True, is a solution

( 2 , 1 ) 2 ( 2 ) + 3 ( 1 ) ≥ 5

7 ≥ 5 True, is a solution

24

Linear Inequalities in Two Variables 2.6

Graph of a linear inequality in two variables is a half-plane. To graph a linear inequality follow these steps:

Step 1: Graph the boundary line of the inequality. Remember to use a dash line for < or > and a solid line for ≤ or ≥

Step 2: Test a point out that is NOT on the line to see what side to shade.

Graph: y < − 2 Graph: x < 1

y < − 2

y = − 2 x

y y

x

x = 1

x < 1

25

Linear Inequalities in Two Variables 2.6

x y < 2 x y

0 y < 2 ( 0 ) 0

1 y < 2 ( 1 ) 2

••

y < 2 x

x 2 x – 5 y ≥ 10 y

0 2 ( 0 ) – 5 y ≥ 10 – 5 – 5 y ≤ −2

−2

5 2 x – 5 ( 0 ) ≥ 10 2 2 x ≥ 5

0•

•2 x – 5 y = 10

Test Point( 1, 1 )

•

Test Point( 0, 0 )

•

26

2 x – 5 y > 10

Graphing Linear Inequalities in Two Variables

y =

2 x

2.7

Piecewise Functions: A combination of equations, each corresponding to part of a domain

Example 1: Evaluating a piecewise function

Evaluate f (x) when x = 0, when x = 2 and when x = 4

Evaluate f (x) = x + 2 , if x < 2

2 x + 1 , if x ≥ 2

Solution:

When x = 0 , f (x) = x + 2

f (0) = 0 + 2 = 2

When x = 2 , f (x) = 2 (x) + 1

f (2) = 2 (2) + 1 = 5

When x = 4 , f (x) = 2 (x) + 1

f (4) = 2 (4) + 1 = 9

{

20•○

27

Piecewise Functions 2.7

Graphing a Piecewise Functions:

f (x) = 1 x + 3 , if x < 1 2 2

− x + 3 , if x ≥ 1{

f (x) = 1 x + 3 2 2

f (1) = 1 (1 ) + 3 = 4 = 2 2 2 2

f ( −3) = 1 ( −3) + 3 = 0 2 2

f ( x ) = − x + 3

f ( 1 ) = − ( 1 ) + 3 = 2

f (3 ) = − (3) + 3 = 0

•

•

•

•

( 1 , 2 )

2 rays with a common initial point, ( 1, 2 )

28

Graphing Piecewise Functions 2.7

Graph the following piecewise function, show all work, label graph

f (x) = 2x – 1 , if x ≤ 13 x + 1 , if x > 1{

x y If x ≤ 1,f (x) = 2 x – 1

If x > 1,f (x) = 3 x + 1

x y

1 1 f (x) = 2 (1) – 1f (x) = 2 – 1

f (x) = 1

f (x) = 3 (1) + 1f (x) = 3 + 1

f (x) = 4

1 4

0 –1 f (x) = 2 (0) – 1f (x) = 0 – 1

f (x) = –1

f (x) = 3 (2) + 1f (x) = 6 + 1

f (x) = 7

2 7

•

•

•

○

x

y

29

Graphing Piecewise Functions 2.7

f (x) = 1 , if 0 ≤ x < 1 2 , if 1 ≤ x < 2 3 , if 2 ≤ x < 3 4 , if 3 ≤ x < 4{

y

x

•

•

•

•

○

○

○

○

It is called a Step function because its graph looks like a set of stair steps

1 2 3 4

1

2

3

4

30

Graphing Piecewise Functions 2.7

• ••○

( 0 , 2 )( 2 , 2 )

(−2 , 0 )

( 0 , 0 )

m = 0 – 2 = − 2 = 1 − 2 – 0 − 2

y = 1 x + 2

m = 0 – 2 = − 2 = 1 0 – 2 − 2

y = 1 x + 0

(−2 , 0 ) , ( 0 , 2 )

(0, 0 ) , ( 2 , 2 )

31

2.7Writing a Piecewise Function

x , if x > 0Remember: │x│ = 0, if x = 0

− x, if x < 0{•

••y = x

x

y

( − 2 , 2 ) ( 2 , 2 )

Vertex

Slope – Intercept Formy = m x + b

Absolute Value Function:

y = a │ x – h │ + k

y = │x │

32

Writing a Piecewise Function y = − x

Absolute Value Function: the a value gives youface up or down and steepness

y = a │ x – h │ + k If a is + , then the graph is face up or

If a is − , then the graph is face down

If a is bigger, then slope is steeper

If a is smaller, then slope is wider

33

Absolute Value Functions 2.8

Absolute Value Function: the h value gives youRight (+) or left (−) movement of vertex

y = a │ x – h │ + k

The more h is + , the more the vertex moves to the right

The more h is − , the more the vertex moves to the left

y = a │ x – ( − 3 ) │ + k y = a │ x – ( 0 ) │ + k y = a │ x – ( 3 ) │ + k

34

Absolute Value Functions 2.8

Absolute Value Function: the h value also gives youLine of Symmetry

y = a │ x – h │ + k

y = a │ x – ( − 3 ) │ + k y = a │ x – ( 0 ) │ + k y = a │ x – ( 3 ) │ + k

x = − 3 x = − 3x = 0 35

Absolute Value Functions

The line of symmetry is in the line x = h

2.8

Absolute Value Function: the k value gives youUp or down movement of the vertex

y = a │ x – h │ + k

The more k is + , the more the vertex moves up

The more k is − , the more the vertex moves down

y = a │ x – h │ + 3 y = a │ x – h │ + 0 y = a │ x – h│ − 3

36

Absolute Value Functions 2.8

(h, k ) values gives you the Vertex of the Absolute Value Function

y = a │ x – h │ + k

•(h, k )

37

Absolute Value Functions 2.8