chapter 2 – linear equations algebra ii. table of contents 2.5 - linear inequalities in two...

Chapter 2 – Linear Equations

Algebra II

Table of Contents

• 2.5 - Linear Inequalities in Two Variables• 2.6 - Transforming Linear Functions• 2.7 - Curve Fitting with Linear Models• 2.8 – Solving Absolute Value Equations and

Inequalities• 2.9 - Absolute Value Function

2.5 - Linear Inequalities in Two Variables

Algebra II

Linear functions form the basis of linear inequalities.

A linear inequality in two variables relates two variables using an inequality symbol, such as y > 2x – 4.

Its graph is a region of the coordinate plane bounded by a line.

The line is a boundary line, which divides the coordinate plane into two regions.

2.5 Algebra 2 (bell work)

For example, the line y = 2x – 4, shown at right, divides the coordinate plane into two parts: one where y > 2x – 4one where y < 2x – 4.

In the coordinate plane higher points have larger y values, so the region where y > 2x – 4 is above the boundary line

where y = 2x – 4.

2.5

Graph the inequality

The boundary line is

which has a y-intercept of 2 and a slope of .

Draw the boundary line dashed because it is not part of the solution.

Then shade the region above the boundary line to show

.

2.5 Example 1 Graphing linear Inequalities

Check Choose a point in the solution region, such as (3, 2) and test it in the inequality.

?

2 > 1 ?

2.5

Graph the inequality y ≤ –1.

Recall that y= –1 is a horizontal line.

Step 1 Draw a solid line for y=–1 because the boundary line is part of the graph.

Step 2 Shade the region below the boundary line to show where y < –1.

2.5

The boundary line is y = 3x – 2 which

has a y–intercept of –2 and a slope of 3.

Draw a solid line because it is part of the solution.

Then shade the region above the boundary line to show y > 3x – 2.

Graph the inequality y ≥ 3x –2.2.5

Check Choose a point in the solution region, such as (–3, 2) and test it in the inequality.

y ≥ 3x –2

2 ≥ 3(–3) –2?

2 ≥ (–9) –2?

2 > –11 ?

2.5

Graph 3x + 4y ≤ 12 using intercepts.

Step 1 Find the intercepts.

Substitute x = 0 and y = 0 into 3x + 4y = 12 to find the intercepts of the boundary line.

y-intercept x-intercept

3x + 4y = 12

3(0) + 4y = 12 3x + 4(0) = 12

4y = 12

3x + 4y = 12

y = 3

3x = 12

x = 4

2.5 Example 2 Graphing Linear Inequalities Using Intercepts

Step 2 Draw the boundary line.The line goes through (0, 3) and (4, 0).

Draw a solid line for the boundary line because it is part of the graph.

Step 3 Find the correct region to shade.Substitute (0, 0) into the inequality.

Because 0 + 0 ≤ 12 is true, shade the region that contains (0, 0).

(0, 3)

(4, 0)

2.5

Graph 3x – 4y > 12 using intercepts.

Step 1 Find the intercepts.

Substitute x = 0 and y = 0 into 3x – 4y = 12 to find the intercepts of the boundary line.

y-intercept x-intercept

3x – 4y = 12

3(0) – 4y = 12 3x – 4(0) = 12

– 4y = 12

3x – 4y = 12

y = – 3

3x = 12

x = 4

2.5

Step 2 Draw the boundary line.The line goes through (0, –3) and (4, 0). Draw the boundary line dashed because it is not part of the solution.

Step 3 Find the correct region to shade.Substitute (0, 0) into the inequality.

Because 0 + 0 >12 is false, shade the region that does not contain (0, 0).

(4, 0)

(0, –3)

2.5

A school carnival charges $4.50 for adults and $3.00 for children. The school needs to make at least $135 to cover expenses.

A. Using x as the number of adult tickets and y as the number of child tickets, write and graph an inequality for the amount the school makes on ticket sales.

B. If 25 child tickets are sold, how many adult tickets must be sold to cover expenses?

An inequality that models the problem is 4.5x + 3y ≥ 135.

135y3.00+x4.50

total.is at least

number of child tickets

timeschild priceplus

number of adult

ticketstimesAdult

price

• •

2.5 Algebra 2 (Bell work)Example 3 Problem-Solving Application

Day 2

Find the intercepts of the boundary line.

Graph the boundary line through

(0, 45) and (30, 0) as a solid line.

Shade the region above the line that is in the first quadrant, as ticket sales cannot be negative.

4.5(0) + 3y = 135 4.5x + 3(0) = 135

y = 45 x = 30

2.5

If 25 child tickets are sold

At least 14 adult tickets must be sold.

4.5x + 3(25) ≥ 135

4.5x + 75 ≥ 135

4.5x ≥ 60

x ≥ 13.3

2.5 B. If 25 child tickets are sold, how many adult tickets must be sold to cover expenses?

A café gives away prizes. A large prize costs the café $125, and the small prize costs $40.

The café will not spend more than $1500.

a) How many of each prize can be awarded?

b) How many small prizes can be awarded if 4 large prizes are given away?

An inequality that models the problem is 40x + 125y ≤ 135.

1500y125+x40

total.is less than

number awardedtimeslarge

prizeplusnumber awardedtimesSmall

prize

≤

2.5 Optional

Find the intercepts of the boundary line.

Graph the boundary line through (0, 12) and (37.5, 0) as a solid line.

Shade the region below the line that is in the first quadrant, as prizes awarded cannot be negative.

40(0) + 125y = 1500 40x + 125(0) = 1500

y = 12 x = 37.5

2.5

No more than 25 small prizes can be awarded.

40x + 125(4) ≤ 1500

40x + 500 ≤ 1500

40x ≥ 1000, so x ≤ 25

2.5 b) How many small prizes can be awarded if 4 large prizes are given away?

You can graph a linear inequality that is solved for y with a graphing calculator.

Press and use the left arrow key to move to the left side.

Each time you press you will see one of the graph styles shown here. You are already familiar with the line style.

2.5

Solve for y. Graph the solution.

8x – 2y > 8

–2y > –8x + 8

y < 4x – 4

2.5

Note that the graph is shown in the STANDARD SQUARE window. (6:ZStandard followed by 5:ZSquare).

Use the calculator option to shade below the line y < 4x – 4.

Solve 2(3x – 4y) > 24 for y. Graph the solution.

–4y > –3x + 12

3x – 4y > 12

Use the calculator option to shade below the line

2.5

HW pg. 128• 2.5

– Day 1: 2-8, 28, 32, 57-64– Day 2: 9, 10, 19-21 (No Graph), 34, 44, 46– Ch: 42

– Follow All HW Guidelines or ½ off

2.6 - Transforming Linear Functions

Algebra II

• Copy the rules below• 1) Horizontal shift f(x) = f(x-h)

h > 0, moves right, f( x – 3)h < 0, moves left, f( x + 5)

• 2) Vertical Shift f(x) = f(x) + kk > 0, moves up, f(x) + 2k < 0, moves down, f(x) - 6


• 3) Reflection across y – axisf(x) = f(-x)

• 4) Reflection across x – axisf(x) = - f(x)

2.6

Let g(x) be the indicated transformation of f(x). Write the rule for g(x).

f(x) = x – 2 , horizontal translation right 3 units

Translating f(x) 3 units right subtracts 3 from each input value.

Subtract 3 from the input of f(x).

Simplify.

g(x) = f(x – 3)

g(x) = (x – 3) – 2

g(x) = x – 5

Evaluate f at x – 3.

2.6 Example 1 Translating and Reflection Linear Functions

Check Graph f(x) and g(x) on a graphing calculator.

The slopes are the same, but the x-intercept has moved 3 units right from 2 to 5.

2.6

linear function defined in the table; reflection across x-axis

x –2 0 2

f(x) 0 1 2


The y-intercept is 1.

Find the slope:

y = mx + b

2.6

Step 2 Write the rule for g(x). Reflecting f(x) across the x-axis replaces each y with –y.

g(x) = –f(x)

2.6

linear function defined in the table; reflection across x-axis

x –2 0 2

f(x) 0 1 2


5) Horizontal Stretch/Compression f(x) = f( x), b > 1 stretches away from the y – axis

0 < │ b │ < 1 compresses toward the y –axis

6) Vertical Stretch/Compression

f(x) = a · f(x), a > 1 stretches away from the x- axis 0 < │ a│ < 1, compresses toward the x - axis

2.6

Math Joke

• Teacher: Why are all your transformations in French?

• Student: They’re translations

2.6

Let g(x) be a horizontal compression of f(x) = –x + 4 by a factor of .

Write the rule for g(x), and graph the function.

Horizontally compressing f(x) by a factor of replaces each x with where b =

g(x) = –2x +4

= –(2x) +4

2.6 Example 2 Stretching and Compressing Linear Functions

Check Graph both functions on the same coordinate plane.

The graph of g(x) is steeper than f(x), which indicates that g(x) has been horizontally compressed from f(x), or pushed toward the y-axis.

2.6

Let g(x) be a vertical compression of f(x) = 3x + 2 by a factor of

Write the rule for g(x) and graph the function.Vertically compressing f(x) by a factor of replaces each f(x) with a · f(x), where a =

g(x) = a(3x + 2)

= (3x + 2)

2.6

Graph both functions on the same coordinate plane.

The graph of g(x) is less steep than f(x),

which indicates that f(x) has been vertically compressed from g(x), or compressed towards the x-axis.

2.6

Day 22.6

Let g(x) be a horizontal shift of f(x) = 3x, left 6 units followed by a horizontal stretch by a factor of 4.

Write the rule for g(x).

Step 1 First perform the translation.

h(x) = f(x + 6)

h(x) = 3(x + 6)

h(x) = 3x + 18

2.6

Step 2 Then perform the stretch.

Stretching h(x) horizontally by a factor of 4

replaces each x with where b = 4.

Example 3 Combining Transformations of Linear Functions

Step 1 First perform the translation.

Translating f(x) = 3x left 8 units adds 8 to each input value.

You can use h(x) to represent the translated function.

Let g(x) be a vertical compression of f(x) = x by a factor of followed by a horizontal

shift 8 left units. Write the rule for g(x).

h(x) = f(x + 8)

h(x) = x + 8

h(x) = x + 8

2.6

Step 2 Then perform the stretch.

Stretching h(x) vertically by a factor of

multiplies the function by .

The golf team is selling T-shirts as a fund-raiser. The function R(n) = 7.5n represents the team’s revenue in dollars, and n is the number of t-shirts sold.

The team paid $60 for the T-shirts. Write a new function P(n) for the team’s profit.

The initial costs must be subtracted from the revenue.

R(n) = 7.5n

P(n) = 7.5n – 60

Graph both functions.

The lines have the same slope but different y-intercepts.

Note that the profit can be negative but the number of T-shirts sold cannot be less than 0.

R

P

2.6 Example 4 Fund Raising Application

The Dance Club is selling beaded purses as a fund-raiser.

The function R(n) = 12.5n, represents the club’s revenue in dollars where n is the number of purses sold.

The club paid $75 for the materials needed to make the purses.

Write a new function P(n) for the club’s profit.

The initial costs must be subtracted from the revenue.

Subtract the expenses. S(n) = 25n – 75

2) What if …? The club members decided to double the price of each purse

2.6

P(n) = 12.5n - 75

Optional

Graph both S(n) and P(n) on the same coordinate plane.

Graph both functions.

The lines have the same slope but different y-intercepts.

Note that the profit can be negative but the number of purses sold cannot be less than 0.

SP

2.6

HW pg. 138

• 2.6-– Day 1: 1-4, 11, 12, 16, 31-38– Day 2: 5-7, 15, 21, 22, 24-27


2.7 – Curve Fitting with Linear Models

Algebra II

The statistical study of the relationship between variables is called regression.

A scatter plot is helpful in understanding the form, direction, and strength of the relationship between two variables.

Correlation is the strength and direction of the linear relationship between the two variables.

Curve Fitting

2.7 Algebra 2 (bell Work)

1. Summarize Definitions Below2. Copy graphs below as well

http://phet.colorado.edu/sims/curve-fitting/curve-fitting_en.html

Albany and Sydney are about the same distance from the equator.

Make a scatter plot with Albany’s temperature as the independent variable.

Name the type of correlation.

Then sketch a line of best fit and find its equation.

2.7 Application By Hand

••••••••••

•

Step 1 Plot the data points.

Step 2 Identify the correlation.

Notice that the data set is negatively correlated–as the temperature rises in Albany, it falls in Sydney.

2.7

Step 3 Sketch a line of best fit.

Draw a line that splits the data evenly above and below.

••••••••••

•

2.7

Step 4 Identify two points on the line.

For this data, you might select (35, 64) and (85, 41).

Step 5 Find the slope of the line that models the data.

Use the point-slope form.

An equation that models the data is y = –0.46x + 80.1.

y – y1= m(x – x1)

y – 64 = –0.46(x – 35)

y = –0.46x + 80.1

2.7

The correlation coefficient r is a measure of how well the data set is fit by a model.

2.7

Math Joke

• Q: why did all the apples in the fruit bowl know each other?

• A: They were core-relations

2.7

To display the correlation coefficient r, you may have to turn on the diagnostic mode. To do this, press and choose the DiagnosticOn mode.

You can use a graphing calculator to perform a linear regression and find the correlation coefficient r.

2.7

Anthropologists can use the femur, or thighbone, to estimate the height of a human being.

The table shows the results of a randomly selected sample.

2.7 Use CalcApplication

••••

• • •

•

a. Make a scatter plot of the data with femur length as the independent variable.

The scatter plot is shown at right.

2.7 Use Calc

b. Find the correlation coefficient r and the line of best fit. Interpret the slope of the line of best fit in the context of the problem.

Enter the data into lists L1 and L2 on a graphing calculator. Use the linear regression feature by pressing

STAT, choosing CALC, and selecting 4:LinReg. The equation of the line of best fit is h ≈ 2.91l + 54.04.

2.7 Use Calc

c. A man’s femur is 41 cm long. Predict the man’s height.

The height of a man with a 41-cm-long femur would be about 173 cm.

h ≈ 2.91(41) + 54.04

The equation of the line of best fit is h ≈ 2.91l + 54.04.

Use the equation to predict the man’s height. For a 41-cm-long femur,

h ≈ 173.35

2.7 Use Calc

The gas mileage for randomly selected cars based upon engine horsepower is given in the table.

1. Make a scatter plot of the data2. What kind of correlation is there3. What is the correlation coefficient (“r value”)?4. What is the equation of the line of best fit?5. Predict the gas mileage for a 210 horsepower engine

Application2.7 Use Calc

•••••

•••••

a. Make a scatter plot of the data with horsepower as the independent variable.

The scatter plot is shown on the right.

2.7 Use Calc

b. Find the correlation coefficient r and the line of best fit. Interpret the slope of the line of best fit in the context of the problem.

Enter the data into lists L1 and L2 on a graphing calculator.

Use the linear regression feature by pressing STAT, choosing CALC, and selecting 4:LinReg.

The equation of the line of best fit is y ≈ –0.15x + 47.5.

2.7 Use Calc

The correlation coefficient is r ≈ –0.916, which indicates a strong negative correlation.

The slope is about –0.15, so for each 1 unit increase in horsepower, gas mileage drops ≈ 0.15 mi/gal.

c. Predict the gas mileage for a 210-horsepowerengine.

Substitute 210 for x.

The mileage for a 210-horsepower engine would be about 16.0 mi/gal.

y ≈ –0.15(210) + 47.50.

The equation of the line of best fit is y ≈ –0.15x + 47.5. Use the equation to predict the gas mileage. For a 210-horsepower engine,

y ≈ 16

2.7 Use Calc

Find the following for this data on average temperature and rainfall for eight months in Boston, MA.

Application2.7 Use Calc

o

•

•

•

••

•

••

a. Make a scatter plot of the data with temperature as the independent variable.


2.7

o

•

•

•

••

•

••

b. Find the correlation coefficient and the equation of the line of best fit. Draw the line of best fit on your scatter plot.

The correlation coefficient is r = –0.703.

The equation of the line of best fit is y ≈ –0.35x + 106.4.

2.7

c. Predict the temperature when the rainfall is 86 mm. How accurate do you think your prediction is?

Rainfall is the dependent variable.

The line predicts 58.3F, but the scatter plot and the value of r show that temperature by itself is not an accurate predictor of rainfall.

86 ≈ –0.35x + 106.4

–20.4 ≈ –0.35x

58.3 ≈ x

2.7

Use the equation of the line of best fit to predict the number of grams of fat in a sandwich with 420 Calories. How close is your answer to the value given in the table?

Find the following information for this data set on the number of grams of fat and the number of calories in sandwiches served at Dave’s Deli.

2.7

a. Make a scatter plot of the data with fat as the independent variable.


2.7

b. Find the correlation coefficient and the equation of the line of best fit. Draw the line of best fit on your scatter plot.

The correlation coefficient is r = 0.682. The equation of the line of best fit is y ≈ 11.1x + 309.8.

2.7

c. Predict the amount of fat in a sandwich with 420 Calories. How accurate do you think your prediction is?

420 ≈ 11.1x + 309.8 Calories is the dependent variable.

110.2 ≈ 11.1x

9.9 ≈ x

The line predicts 10 grams of fat. This is not close to the 15 g in the table.

2.7

HW pg. 146

• 2.7-– 2, 3 (By Hand, Estimate “r”) – 4, 7, 15 (Use Calc)– 16, 18, 25-31


2.8 – Solving Absolute – Value Equations and Inequalities

Algebra II

A compound statement is made up of more than one equation or inequality.

A disjunction is a compound statement that uses the word or.

Disjunction: x ≤ –3 OR x > 2 Set builder notation: {x|x ≤ –3 U x > 2} “U for Union”

A disjunction is true if and only if at least one of its parts is true.


A conjunction is a compound statement that uses the word and.

Conjunction: x ≥ –3 AND x < 2 Set builder notation: {x|x ≥ –3 x < 2}.

“n for intersection”

A conjunction is true if and only if all of its parts are true. Conjunctions can be written as a single statement as shown.

x ≥ –3 and x< 2 –3 ≤ x < 2U

2.8

Solve the compound inequality. Then graph the solution set.

Solve both inequalities for y.

The solution set is all points that satisfy {y|y < –4 or y ≥ –2}.

6y < –24 OR y +5 ≥ 3

6y < –24 y + 5 ≥3

y < –4 y ≥ –2

or

–6 –5 –4 –3 –2 –1 0 1 2 3

(–∞, –4) U [–2, ∞)

2.8 Example 1 Solving Compound Inequalities

Solve both inequalities for c.

The solution set is the set of points that satisfy both c ≥ –4 and c < 0.

c ≥ –4 c < 0

–6 –5 –4 –3 –2 –1 0 1 2 3

[–4, 0)

and 2c + 1 < 1


Solve both inequalities for x.

The solution set is the set of all points that satisfy {x|x < 3 or x ≥ 5}.

–3 –2 –1 0 1 2 3 4 5 6

x – 5 < –2 OR –2x ≤ –10

x < 3 x ≥ 5

x – 5 < –2 or –2x ≤ –10

(–∞, 3) U [5, ∞)

2.8 Optional

2.8Do not Copy

Absolute-value equations and inequalities can be represented by compound statements. Consider the equation |x| = 3.

The solutions of |x| = 3 are the two points that are 3 units from zero. The solution is a disjunction: x = –3 or x = 3.

2.8

The solutions of |x| < 3 are the points that are less than 3 units from zero. The solution is a conjunction: –3 < x < 3.

2.8

The solutions of |x| > 3 are the points that are more than 3 units from zero. The solution is a disjunction: x < –3 or x > 3.

2.8

Math Joke

• Q: Why are you so negative?

• A: Just take me for my absolute value!

2.8

Note: The symbol ≤ can replace <, and the rules still apply.

The symbol ≥ can replace >, and the rules still apply.

2.8

Solve the equation.

x = 16 or x = –16

Solve the equation.

|–3 + k| = 10

–3 + k = 10 or –3 + k = –10

k = 13 or k = –7

2.8 Example 2 Solving Absolute-Value Equations

|x + 9| = 13

Solve the equation.

x + 9 = 13 or x + 9 = –13

x = 4 or x = –22

|6x| – 8 = 22

Solve the equation.

|6x| = 30

6x = 30 or 6x = –30

x = 5 or x = –5

2.8 Optional

You can solve absolute-value inequalities using the same methods that are used to solve an absolute-value equation.


Solve the inequality. Then graph the solution.|–4q + 2| ≥ 10

–4q + 2 ≥ 10 or –4q + 2 ≤ –10

–4q ≥ 8 or –4q ≤ –12

q ≤ –2 or q ≥ 3

2.8

1. Drop the absolute value bars and solve for first case.

2. Then Flip inequality sign and change sign on constant for second case

Example 3 Solving Absolute-Value Inequalities with Disjunctions

–3 –2 –1 0 1 2 3 4 5 6

(–∞, –2] U [3, ∞)

To check, you can test a point in each of the three region.

|–4(–3) + 2| ≥ 10

|14| ≥ 10

|–4(0) + 2| ≥ 10

|2| ≥ 10 x

|–4(4) + 2| ≥ 10

|–14| ≥ 10

2.8

Solve the inequality. Then graph the solution.

|0.5r| – 3 ≥ –3

0.5r ≥ 0 or 0.5r ≤ 0

r ≤ 0 or r ≥ 0

|0.5r| ≥ 0

–3 –2 –1 0 1 2 3 4 5 6

(–∞, ∞)

The solution is all real numbers, R.

2.8


|p – 2| ≤ –6

|p – 2| ≤ –6 and p – 2 ≥ 6

p ≤ –4 and p ≥ 8

Because no real number satisfies both p ≤ –4 and p ≥ 8, there is no solution.

The solution set is ø.

2.8 Example 4 Solving Absolute-Value Inequalities with Conjunctions

Solve the inequality. Then graph the solution.

|4x – 8| > 12

4x – 8 > 12 or 4x – 8 < –12

4x > 20 or 4x < –4

x > 5 or x < –1

–3 –2 –1 0 1 2 3 4 5 6

(–∞, ∞)The solution is all real numbers, R.

2.8

|x – 5| ≤ 8

x – 5 ≤ 8 and x – 5 ≥ –8

x ≤ 13 and x ≥ –3

The solution set is {x|–3 ≤ x ≤ 13}.

–10 –5 0 5 10 15 20 25

2.8

HW pg. 154

• 2.8-– Day 1: 2-7, 14, 15, 19, 29, 31, – Day 2: 8-13, 26, 27, 44, 71– Ch: 54


2.9 – Absolute Value Function

Algebra II

An absolute-value function is a function whose rule contains an absolute-value expression.

The graph of the parent absolute-value function f(x) = |x| has a V shape with a minimum point or vertex at (0, 0).

2.9 Algebra 2 (bell work)1. Summarize the definition and copy the graph below

Examples of Absolute value in real life• Distance

• Years AD, BC

• Model weights such as length/size of baseball bats

• Depth below ocean

• Speed

• Computer Programming, without there would be no social networking (a good thing?)

• Money and finances

2.9

The general forms for translations are

Vertical:

g(x) = f(x) + k

Horizontal:

g(x) = f(x – h)

Remember!

Reflection across x-axis: g(x) = –f(x)

Reflection across y-axis: g(x) = f(–x)

Remember!

Vertical stretch and compression : g(x) = af(x)

Horizontal stretch and compression: g(x) = f

Remember!

2.9

Perform the transformation on f(x) = |x|. Then graph the transformed function g(x).

5 units down

f(x) = |x|

g(x) = f(x) + k

g(x) = |x| – 5

f(x)

g(x)

2.9 Example 1 Translating Absolute-Value Functions

Perform the transformation on f(x) = |x|. Then graph the transformed function g(x).

1 unit left

f(x) = |x|

g(x) = f(x – h )

g(x) = |x – (–1)| = |x + 1|

f(x)

g(x)

2.9

Perform the transformation on f(x) = |x|. Then graph the transformed function g(x). 2 units right

f(x) = |x|

g(x) = f(x – h)

g(x) = |x – 2| = |x – 2|

f(x)

g(x)

2.9

4 units up

f(x) = |x|

g(x) = f(x) + k

g(x) = |x| + 4

Let g(x) be the indicated transformation of f(x) = |x|. Write the rule for g(x) and graph the function.

f(x)

g(x)

2.9

Math Joke

• Q: Why was the parent function upset with its child?

• A: It was stretched to the limit

2.9

Because the entire graph moves when shifted, the shift from f(x) = |x| determines the vertex of an absolute-value graph.

2.9

Translate f(x) = |x| so that the vertex is at (–1, –3). Then graph.

g(x) = |x – h| + k

g(x) = |x – (–1)| + (–3)

g(x) = |x + 1| – 3f(x)

g(x)

2.9 Translations of an Absolute-Value FunctionExample 2

Translate f(x) = |x| so that the vertex is at (4, –2). Then graph.

g(x) = |x – h| + k

g(x) = |x – 4| + (–2)

g(x) = |x – 4| – 2g(x)

f(x)

2.9

Perform the transformation. Then graph.

g(x) = f(–x)

g(x) = |(–x) – 2| + 3

Reflect the graph. f(x) =|x – 2| + 3 across the y-axis.

gf

2.9 Example 3 Transforming Absolute-Value Functions Day 2

g(x) = af(x)

g(x) = 2(|x| – 1)

Stretch the graph. f(x) = |x| – 1 vertically by a factor of 2.

g(x) = 2|x| – 2

f(x) g(x)

2.9

Compress the graph of f(x) = |x + 2| – 1 horizontally by a factor of

g(x) = |2x + 2| – 1

f

g

2.9

Stretch the graph. f(x) = |4x| – 3 horizontally by a factor of 2.

g(x) = |2x| – 3

g(x) = f( x)

g(x) = | (4x)| – 3 g f

2.9

Perform the transformation. Then graph.

g(x) = f(–x)

g(x) = –|–x – 4| + 3

Reflect the graph. f(x) = –|x – 4| + 3 across the y-axis.

g(x) = –|(–x) – 4| + 3

fg

2.9

HW pg. 161

• 2.9-– Day 1: 2-5, 9-14, 49-53 (Odd)– Day 2: 6-8, 15-17, 21, 55, 56– Ch: 32


chapter 2 – linear equations algebra ii. table of contents 2.5 - linear inequalities in two...

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