Chapter 2 - Linear Functions Algebra II

Table of Contents 2.1- Solving Linear Equations and Inequalities 2.2- Proportional Reasoning 2.3- Graphing Linear Functions 2.4- Writing Linear Functions

2.1- Solving Linear Equations and Inequalities Algebra II

2-1 Algebra 2 (bell work) An equation is a mathematical statement that two expressions are equivalent The solution set of an equation is the value or values of the variable that make the equation true. A liner equation is one variable can be written in the form ax=b, where a and b are constants and a ≠ 0

Let m represent the number of additional minutes that Nina used. 2-1 Example 1 Application The local phone company charges $12.95 a month for the first 200 of air time, plus $0.07 for each additional minute. If Nina’s bill for the month was $14.56, how many additional minutes did she use? Let m represent the number of additional minutes that Nina used. Model additional minute charge number of additional minutes monthly charge total charge plus times = = 14.56 12.95 + 0.07 * m

2-1 The local phone company charges $12.95 a month for the first 200 of air time, plus $0.07 for each additional minute. If Nina’s bill for the month was $14.56, how many additional minutes did she use? Let m represent the number of additional minutes that Nina used. Solve 12.95 + 0.07m = 14.56 –12.95 - 12.95 0.07m = 1.61 0.07 0.07 m = 23 Nina used 23 additional minutes.

number of additional cups 2-1 Stacked cups are to be placed in a pantry. One cup is 3.25 in. high and each additional cup raises the stack 0.25 in. How many cups fit between two shelves 14 in. apart? additional cup height number of additional cups one cup total height plus times = 3.25 = 14.00 + 0.25 * c

Stacked cups are to be placed in a pantry. 2-1 Stacked cups are to be placed in a pantry. One cup is 3.25 in. high and each additional cup raises the stack 0.25 in. How many cups fit between two shelves 14 in. apart? Solve 3.25 + 0.25c = 14.00 –3.25 –3.25 0.25c = 10.75 0.25 c = 43 44 cups fit between the 14 in. shelves.

2-1 Example 2 Solving Equations with the Distributive Property Solve 4(m + 12) = –36 Solve 4(m + 12) = –36 Method 1 Method 2 The quantity (m + 12) is multiplied by 4, so divide by 4 first. Distribute before solving 4m + 48 = –36 –48 –48 4(m + 12) = –36 4 4 4m = –84 m + 12 = –9 = 4m –84 4 4 –12 –12 m = –21 m = –21

2-1 Solve 3(2 –3p) = 42. Solve 3(2 – 3p) = 42 Method 1 Method 2 6 – 9p = 42 3(2 – 3p) = 42 –6 –6 3 3 –9p = 36 2 – 3p = 14 –2 –2 = –9p 36 –9 –9 –3p = 12 p = –4 –3 –3 p = –4

2-1 Solve –3(5 – 4r) = –9 –15 + 12r = –9 +15 +15 12r = 6 = 12r 6 12 12 r =

Just Read If there are variables on both sides of the equation Simplify each side. Collect all variable terms on one side and all constants terms on the other side. Isolate the variables

2-1 Example 3 Solving Equations with Variables on Both Sides Solve 3k– 14k + 25 = 2 – 6k – 12 Solve 3(w + 7) – 5w = w + 12 –2w + 21 = w + 12 –11k + 25 = –6k – 10 +11k +11k +2w +2w 25 = 5k – 10 21 = 3w + 12 +10 + 10 –12 –12 35 = 5k 9 = 3w 5 5 3 3 7 = k 3 = w

Define the following words 2-1 Algebra 2 (bell work) Define the following words Identity – An equation that is true for all values of the variable, such as x = x Answer is all real numbers, or R, or ( - ∞ , ∞ ) Contradiction – An equation that has no solution, such as 3 = 5. Can also use the empty set symbol

 2-1 Example 4 Identifying Identities and Contradictions Solve 2(x – 6) = –5x – 12 + 7x Solve 3v – 9 – 4v = –(5 + v). 2x – 12 = 2x – 12 3v – 9 – 4v = –(5 + v) –2x –2x –9 – v = –5 – v + v + v –12 = –12  The solutions set is all real number, or . –9 ≠ –5 x The equation has no solution. The solution set is the empty set, which is represented by the symbol .

 2-1 Solve 3(2 –3x) = –7x – 2(x –3) Solve 5(x – 6) = 3x – 18 + 2x 6 = 6 x –30 ≠ –18 The solutions set is all real numbers, or . The equation has no solution. The solution set is the empty set, which is represented by the symbol

2-1 Just Read If you multiply or divide by a negative number, flip the inequality sign These properties also apply to inequalities expressed with >, ≥, and ≤.

2-1 Example 5 Solving Inequalities Solve and graph 8a –2 ≥ 13a + 8 Solve and graph x + 8 ≥ 4x + 17 8a – 2 ≥ 13a + 8 x + 8 ≥ 4x + 17 –13a –13a –x –x –5a – 2 ≥ 8 8 ≥ 3x +17 +2 +2 –17 –17 –5a ≥ 10 –9 ≥ 3x –5a ≤ 10 –9 ≥ 3x 3 3 –5 –5 –3 ≥ x or x ≤ –3 a ≤ –2

HW pg.94 2.1- Day 1: 2-11, 21-25 (Odd), 41, 42, 45, 47 Day 2: 12-20, 34-39, 49, 62-65 Ch: 52 Follow All HW Guidelines or ½ off

2.2- Proportional Reasoning Algebra II

2-2 Copy boxed parts below

2-2 Example 1 Solve each proportion. 14 c = 16 24 p 12.9 A. B. = 88 132 16 24 p 12.9 14 c 88 132 = = 206.4 = 24p 88c = 1848 206.4 24p 24 24 = 88c 1848 = 88 88 8.6 = p c = 21

2-2 Optional Solve each proportion. y 77 12 84 15 2.5 A. = B. = x 7 y 77 12 84 15 2.5 x 7 = = 924 = 84y 2.5x =105 924 84y 84 84 = = 2.5x 105 2.5 2.5 11 = y x = 42

2-2 Because percents can be expressed as ratios, you can use the proportion to solve percent problems. Percent is a ratio that means per hundred. For example: 30% = 0.30 = Remember! 30 100

Percent (as decimal)  whole = part 2-2 Example 2 Solving Percent Problems A poll taken one day before an election showed that 22.5% of voters planned to vote for a certain candidate. If 1800 voters participated in the poll, how many indicated that they planned to vote for that candidate? Method 1 Use a proportion. Method 2 Use a percent equation. Percent (as decimal)  whole = part 0.225  1800 = x 405 = x x = 405 So 405 voters are planning to vote for that candidate.

Percent (as decimal)  whole = part 2-2 At Clay High School, 434 students, or 35% of the students, play a sport. How many students does Clay High School have? Method 1 Use a proportion. Method 2 Use a percent equation. 35% = 0.35 Percent (as decimal)  whole = part 0.35x = 434 Cross multiply. 100(434) = 35x x = 1240 Solve for x. x = 1240 Clay High School has 1240 students.

600 m 482 strides x m 1 stride  ≈ 2-2 Example 3 Fitness Application (Dimensional Analysis) Ryan ran 600 meters and counted 482 strides. How long is Ryan’s stride in inches? (Hint: 1 m ≈ 39.37 in.) Use a proportion to find the length of his stride in meters. 600 m 482 strides x m 1 stride = 600 = 482x x ≈ 1.24 m  ≈ 1.24 m 1 stride length 39.37 in. 1 m 49 in. Ryan’s stride length is approximately 49 inches.

400 m 297 strides x m 1 stride 2-2 Optional Luis ran 400 meters in 297 strides. Find his stride length in inches. Use a proportion to find the length of his stride in meters. 400 m 297 strides x m 1 stride = 400 = 297x x ≈ 1.35 m  ≈ 1.35 m 1 stride length 39.37 in. 1 m 53 in. Luis’s stride length is approximately 53 inches.

Scaling Geometric Figures in the Coordinate Plane 2-2 Example 4 Scaling Geometric Figures in the Coordinate Plane = height of ∆XAB width of ∆XAB height of ∆XYZ width of ∆XYZ Y Z = 3 x 9 6 B A X 9x = 18, so x = 2

Math Joke Q: Why were the similar triangles weighing themselves A: they were finding their Scale

= Shadow of tree Height of tree Shadow of house Height of house 2-2 The tree in front of Luka’s house casts a 6-foot shadow at the same time as the house casts a 22-foot shadow. If the tree is 9 feet tall, how tall is the house? 9 ft 6 ft = Shadow of tree Height of tree Shadow of house Height of house = 6 9 h 22 h ft 22 ft 6h = 198 h = 33 The house is 33 feet high.

= Shadow of climber Height of climber Shadow of tree Height of tree 2-2 Optional A 6-foot-tall climber casts a 20-foot long shadow at the same time that a tree casts a 90-foot long shadow. How tall is the tree? 6 ft 20 ft = Shadow of climber Height of climber Shadow of tree Height of tree = 20 6 h 90 h ft 90 ft 20h = 540 h = 27 The tree is 27 feet high.

HW pg. 100 2.2- 7, 9- 11, 13, 18, 19, 21, 22, 25, 33, 36, 38, 68-71 Ch: 35 Follow All HW Guidelines or ½ off

Algebra II (Bell work) Copy down the definition of a linear function 2-3 Algebra II (Bell work) Copy down the definition of a linear function Functions with a constant rate of change are called linear functions. A linear function can be written in the form f(x) = mx + b, where x is the independent variable and m and b are constants. (m = slope) (b = y-intercept) The graph of a linear function is a straight line made up of all points that satisfy y = f(x).

2.3- Graphing Linear Functions Algebra II

Determine whether the data set could represent a linear function. 2-3 Example 1 Recognizing Linear Functions Determine whether the data set could represent a linear function. +2 –1 +2 –1 +2 –1 x –2 2 4 f(x) 1 –1 The rate of change, = , is constant So the data set is linear.

Determine whether the data set could represent a linear function. 2-3 Determine whether the data set could represent a linear function. +1 +2 +1 +4 +1 +8 x 2 3 4 5 f(x) 8 16 The rate of change, 2 ≠ 4 ≠ 8, is not constant. So the data set is not linear.

Determine whether the data set could represent a linear function. 2-3 Determine whether the data set could represent a linear function. –2 –2 –4 –2 –8 x 10 8 6 4 f(x) 7 5 1 –7 The rate of change, , is not constant. So the data set is not linear.

2-3 The constant rate of change for a linear function is its slope. The slope of a linear function is the ratio or m= =

Graph the line with slope m= that passes through (–1, –3). 2-3 Example 2 Graphing Lines using Slope and a Point Graph the line with slope m= that passes through (–1, –3).

Graph the line with slope m = that passes through (0, 2). 2-3 Graph the line with slope m = that passes through (0, 2).

Copy the definitions below The y-intercept is the 2-3 Copy the definitions below The y-intercept is the y-coordinate of a point where the line crosses the y-axis. The x-intercept is the x-coordinate of a point where the line crosses the x-axis.

Find the intercepts of 4x – 2y = 16, and graph the line. 2-3 Example 3 Graphing Lines Using the Intercepts Find the intercepts of 4x – 2y = 16, and graph the line. Find the x-intercept: 4x – 2y = 16 4x – 2(0) = 16 x-intercept y-intercept 4x = 16 x = 4 Find the y-intercept: 4x – 2y = 16 4(0) – 2y = 16 –2y = 16 y = –8

Find the intercepts of 6x – 2y = –24, and graph the line. 2-3 Find the intercepts of 6x – 2y = –24, and graph the line. Find the x-intercept: 6x – 2y = –24 6x – 2(0) = –24 x-intercept y-intercept 6x = –24 x = –4 Find the y-intercept: 6x – 2y = –24 6(0) – 2y = –24 –2y = –24 y = 12

2-3 Algebra II (Bell work) Linear functions can also be expressed as linear equations of the form y = mx + b. When a linear function is written in the form y = mx + b The function is said to be in slope-intercept form because m is the slope of the graph and b is the y-intercept. Notice that slope-intercept form is the equation solved for y. So slope-intercept form is: y = mx + b or f(x) = mx + b, where m = slope, b = y-int

Math Joke Q: Why was the student afraid of the y-intercept A: She thought she’d be stung by the b

Solve for y first. –4x + y = –1 +4x +4x y = 4x – 1 2-3 Example 4 Graphing Functions in Slope-Intercept Form Write the function –4x + y = –1 in slope-intercept form. Solve for y first. –4x + y = –1 +4x +4x y = 4x – 1 The line has y-intercept –1 and slope 4, which is . Plot the point (0, –1). Then move up 4 and right 1 to find other points.

Write the function in slope-intercept form. Then graph the function. 2-3 Write the function in slope-intercept form. Then graph the function. Solve for y first. The line has y-intercept 8 and slope . Plot the point (0, 8). Then move down 4 and right 3 to find other points.

Solve for y first. 5x = 15y + 30 –30 –30 5x – 30 = 15y 2-3 Write the function 5x = 15y + 30 in slope-intercept form. Then graph the function. Solve for y first. 5x = 15y + 30 –30 –30 5x – 30 = 15y The line has y-intercept –2 and slope . Plot the point (0, –2). Then move up 1 and right 3 to find other points.

Vertical and Horizontal Lines 2-3 Vertical and Horizontal Lines Vertical Lines Horizontal Lines The line x = a is a vertical line at a. The line y = b is a horizontal line at b. The slope of a vertical line is undefined. The slope of a horizontal line is zero.

Determine if each line is vertical or horizontal. 2-3 Example 5 Graphing Vertical and Horizontal Lines Determine if each line is vertical or horizontal. A. x = 2 This is a vertical line located at the x-value 2. (Note that it is not a function.) x = 2 y = –4 B. y = –4 This is a horizontal line located at the y-value –4.

Determine if each line is vertical or horizontal. 2-3 Determine if each line is vertical or horizontal. A. y = –5 This is a horizontal line located at the y-value –5. x = 0.5 y = –5 B. x = 0.5 This is a vertical line located at the x-value 0.5.

HW pg. 109 2.3- Day 1: 3 – 12, 52, 57, 67-79 Day 2: 13- 21 (21 No Graph), 33, 35, 47, 48, 53, 55ab Follow All HW Guidelines or ½ off Don’t forget to write assignments on hw sheet (or planner)

2.4 Algebra II (Bell work) Copy the boxed parts below

2.4- Writing Linear Functions Algebra II

2.4 Example 1 Writing the Slope-Intercept Form of the Equation of the Line Write the equation of the graphed line in slope-intercept form. Step 1 Identify the y-intercept. The y-intercept b is 1. Step 2 Find the slope. Slope is = = – . Rise run –3 4 3 Step 3 Write the equation in slope-intercept form. 3 –4 4 –3 y = mx + b 3 4 y = – x + 1

2.4 Write the equation of the graphed line in slope-intercept form. Step 1 Identify the y-intercept. The y-intercept b is 3. Slope is = . rise run 3 4 Write the equation in slope-intercept form. y = mx + b 3 4 y = x + 3 m = and b = 3. 3 4 The equation of the line is 3 4 y = x + 3.

2.4 Example 2 Finding the Slope of a Line Given Two or More Points Find the slope of the line through (–1, 1) and (2, –5). Let (x1, y1) be (–1, 1) and (x2, y2) be (2, –5). The slope of the line is –2.

x 4 8 12 16 y 2 5 11 2.4 Find the slope of the line. Let (x1, y1) be (4, 2) and (x2, y2) be (8, 5). The slope of the line is . 3 4

2.4 Find the slope of the line shown. Let (x1, y1) be (0,–2) and (x2, y2) be (1, –2). The slope of the line is 0.

2.4 Because the slope of line is constant, it is possible to use any point on a line and the slope of the line to write an equation of the line in point-slope form.

2.4 Math Joke Student: I’ll just draw a quick line by hand and guess the slope Teacher: No, No. It’s point-slope form, not point-sloppy form

x –8 –4 4 8 y –5 –3.5 –0.5 1 2.4 Example 3 Writing Equations of Lines In slope-intercept form, write the equation of the line that contains the points in the table. x –8 –4 4 8 y –5 –3.5 –0.5 1 First, find the slope. Let (x1, y1) be (–8, –5) and (x2, y2) be (8, 1). Next, choose a point, and use either form of the equation of a line.

2.4 Method A Point-Slope Form Method B Slope-intercept Form Using (8, 1): Using (8, 1), solve for b. y – y1 = m(x – x1) y = mx + b 1 = 3 + b b = –2 Rewrite in slope-intercept form. The equation of the line is .

2.4 Write the equation of the line in slope-intercept form through (–2, –3) and (2, 5). First, find the slope. Let (x1, y1) be (–2,–3) and (x2, y2) be (2, 5). Method A Point-Slope Form Method B Slope-Intercept Form y – y1 = m(x – x1) y = mx + b 5 = (2)2 + b y – (3) = –5(x – 1) 5 = 4 + b y – 3 = –5(x – 1) 1 = b Rewrite in slope-intercept form. Rewrite the equation using m and b. y – 3 = –5(x – 1) y – 3 = –5x + 5 y = 2x + 1 y = mx + b y = –5x + 8 The equation of the line is y = 2x + 1.

2.4 Algebra II (Bell Work) Summarize the boxed parts below

2.4 Write the equation of the line in slope-intercept form. perpendicular to and through (9, –2) parallel to y = 1.8x + 3 and through (5, 2) The slope of the given line is , so the slope of the perpendicular line is the opposite reciprocal, m = 1.8 y – 2 = 1.8(x – 5) y – 2 = 1.8x – 9 y = 1.8x – 7

2.4 Write the equation of the line in slope-intercept form. parallel to y = 5x – 3 and through (1, 4) perpendicular to and through (9, –2) The slope of the given line is , so the slope of the perpendicular line is the opposite reciprocal, . m = 5 y – 4 = 5(x – 1) y – 4 = 5x – 5 y = 5x – 1

Let x = selling price and y = rent. 2.4 Example 4 Entertainment Application The table shows the rents and selling prices of properties from a game. Selling Price ($) Rent ($) 75 9 90 12 160 26 250 44 Express the rent as a function of the selling price. Let x = selling price and y = rent. Find the slope by choosing two points. Let (x1, y1) be (75, 9) and (x2, y2) be (90, 12).

2.4 To find the equation for the rent function, use point-slope form. y – y1 = m(x – x1)

2.4 Graph the relationship between the selling price and the rent. How much is the rent for a property with a selling price of $230? To find the rent for a property, use the graph or substitute its selling price of $230 into the function. y = 46 – 6 y = 40 The rent for the property is $40.

Items Cost ($) 4 14.00 7 21.50 18 2.4 Optional Express the cost as a linear function of the number of items. Let x = items and y = cost. Find the slope by choosing two points. Let (x1, y1) be (4, 14) and (x2, y2) be (7, 21.50).

2.4 To find the equation for the number of items, use point-slope form. y – y1 = m(x – x1) y – 14 = 2.5(x – 4) y = 2.5x + 4

HW pg. 120 2.4 Day 1: 1-8, 17, 27, 43, 52, 54 Day 2: 9-11, 19, 23-25, 38, 44, 49, Ch: 28 Follow All HW Guidelines or ½ off Don’t forget to write assignments on hw sheet (or planner)