1. Introduction Modeling, Functions, and Graphs

2. Modeling The ability to model problems or phenomena by algebraic expressions and equations is the ultimate goal of any algebra course The examples are designed with interactive investigation and to show how mathematical techniques are used for learning  get opportunity to explore open-minded modeling problems.  understand new situation

3. Functions Functions are useful not only in Calculus but in nearby every field students may pursue. We employ celebrated “ Rule of Four” all problems should be considered using Algebraic Method  Verbal  Algebraic Expression  Numerical  Graphical Students learn to write algebraic expression from verbal description, to recognize trends in a table of data, extract & interpret information from the graph of a function

4. Graphs  No tool for conveying information about a system is more powerful than a graph  Large number of examples are explained plotting by hand and using graphing calculator(TI 83/84)

5. Ch 1 - Linear Model Mathematical techniques are used to Analyze data Identify trends Predict the effects of change  This quantitative methods are the concepts of skills of algebra  You will use skills you learned in elementary algebra to solve problems and to study a variety of phenomena  the description of relationships between variables by using equations, graphs, and table of values. This process is called Mathematical Modeling

6. Some examples of Linear Models ( Example 1 pg – 3) 0 1 2 3 4 5 Length of Rentals 20 15 10 5 Length of Cost of Rental rental (dollars) (hours) 0 5 1 8 2 11 3 14 (t, C) (0, 5) (1, 8) (2, 11) (3, 14) C = 5 + 3(0) C = 5 + 3(1) C = 5 + 3(2) C = 5 + 3(3) Equation C = ($ 5 isurance fee) + $ 3 x (number of hours) C = 5 + 3. t Cost of Rentals

7. Exercise 1.1 ( Example 2, pg – 10) 0 4 8 12 16 20 24 200 180 175 150 125 100 75 50 25 The Equation g = 20 – 1 m 12 m 0 48 96 144 192 g 20 16 12 8 4 Let g = 5, 5 = 20 – 1 m 12 60 = 240 – m (Multiply by 12 both sides ) - - 180 = - m (multiply by – 1 both sides) m = 180 Leon has traveled more than 180 miles if he has less than 5 gallons of gas left 4 gallons m g Table

8. Intercepts Consider the graph of the equation 3y – 4x = 12 The y-coordinate of the x-intercept is zero, so we set y = 0 in the equation to get 3(0) – 4x = 12 x = - 3 The x-intercept is the point (-3, 0). Also, the x-coordinate of the y-intercept is zero, so we set x = 0 in the equation to get 3y – 4(0) = 12 y = 4 The y-intercept is (0, 4) ( -3, 0) (0, 4)

9. Example 9, Pg = 10 x - y = 1 9 4 Set x = 0, 0 y 9 – 4 = 1 - y 4 = 1, y = - 4 The y-intercept is the point (0, - 4) Set y = 0, x - 0 = 1 9 4 x = 1, x = 9 9 The x-int is the point (9, 0) (0, -4) (9, 0) Find the intercepts and graph the equation

10. 1.2 Using Graphing Calculator Solving y in terms of x Press Y1=2x - 5 Press 2 nd and Table Press Zoom and 6 Pg 12, 2x – y = 5 Press Y1 = -1.5x + 0.6 Press window and Enter Pg 14, Example 2, 3x + 2y = 16 Press Zoom and 6

11. Finding Coordinates with a Graphing Calculator (Pg 15) Press Y1 enter Press 2 nd and Table Press Graph and then press, Trace and enter “Bug” begins flashing on the display. The coordinates of the bug appear at the bottom of the display.Use the left and right arrows to move the bug along the graph Graph the equation Y = -2.6x – 5.4 X min = -5, X max = 4.4, Y min = - 20, Y max = 15

12. Using Graphing Calculator to solve the equation, (Pg 19) Equation 572 – 23x = 181 Press Window and enter Press Y1 and Y2 and enter Press 2 nd and Table Press Graph and Trace Y1 Y2

13. Graphical Solution of Inequalities (Pg – 17) Consider the inequality 285 – 15x > 150 x 0 2 4 6 8 10 12 285 – 15x 285 255 225 195 165 135 105 10 25 300 200 150 100 - 100 y = 285 – 15x

14. 1.3 Measuring Steepness ( pg 23) Which path is more strenuous ? 5 ft 2 ft Steepness measures how sharply the altitude increases.. To compare the steepness of two inclined paths, we compute the ratio of change in horizontal distance for each path

15. 1.3 Slope (Ex- 3, Pg 25) Definition of Slope: The slope of a line is the ratio Change in y- Coordinate Change in x- coordinate A B 0 2 3 4 5432154321 Slope = Change in y-coordinate = 5 - 4 = 1 Change in x- coordinate 4 – 2 2

16. Notation for Slope (Pg 26) x y y x m =, where x is not equal to zero Slope of a line is given by

17. Significance of the slope (Ex 5, Pg 28) 1 2 3 4 5 t 250 200 150 100 50 t = 2 D = 100 H (4, 200) G (2, 100) D Change in distance = 100miles = 50 miles per hour T Change in time 2 hours Slope m = No of hours Distance in miles traveled The slope represents the trucker’s average speed or velocity

18. 1.4 Equations of Lines ( Pg 40) y = 2x y = ½ x y = -2x These lines have the same y- intercept but different slopes y = 2x + 3 y = 2x + 1 y = 2x - 2 These lines have the same slope but different y - intercepts

19. 1.4 Slope – Intercept Form y = mx + b ( m is slope of a line, and b is the y- intercept) 3x + 4y = 6 Subtract 3x from both sides 4y = - 3x + 6 Divide both sides by 4 y = - 3x + 3 4 2 m= -3 and b = 3 4 2 General Slope- Intercept Method of Graphing b Start here New point y = mx + b x y

20. A Formula for Slope Slope Formula m = y 2 – y 1 = 9-(-6) -15 = - 3 x 2 – x 1 7 – 2 5 m = Y x P 1 (2, 9) P 2 (7, -6) 10 5 -5 10 The slope of the line passing through the points P 1 (x 1, y 1) and P 2 ( x 2, y 2 ) is given by

21. Point- Slope Form Point- Slope Form y – y1 = m (x – x1) x = 4 y = -3 (1, - 4) (5, - 7) 5 - 7 - 4

22. ScatterPlots scattered not organizeddecreasing trendincreasing trend

23. 1.5 Linear Regression (Lines of Best Fit)  The datas in the scatterplot are roughly linear, we can estimate the location of imaginary”lines best fit” that passes as close as possible to the data points  We can make the predictions about the data. The process of predicting a value of y based on a straightline that fits the data is called a linear regression, and the line itself called the regression line.  The equation of the regression line is usually used(instead of graph) to predict values

24. Example of Linear Regression ( pg 55) 1000 2000 Boiling Point 0 C 200 100 Heat of Vaporization (kJ) (900, 100) (1560, 170) a) Slope = m= 170 – 100 = 0.106 1560 – 900 The equation of regression line is y- y1 = m(x – x1) y – 100 = 0.106(x – 900), y = 0.106x + 4.6 b) Regression equation for potassium bromide, x = 1435 y = 0.106(1435) + 4.6 a)Estimate a line of best fit and find the equation of the regression line b)Use the regression line to predict the heat of vaporization of potassium bromide, whose boiling temperature is 1435 0 C Choose two points in the regression line

25. 112 96 80 64 48 20 40 60 80 100 Age (months) The graph is not linear because her rate of growth is not constant; her growth slows down as she approaches her adult height. The short time of interval the graph is close to a line, and that line can be used to approximate the coordinates of points on the curve. Height (cm) Interpolation- The process of estimating between known data points Extrapolation- Making predictions beyond the range of known data

26. Ex 1.5 No 13, Pg 65 Loss in Mass (mg) Vol. Of Gas Cubic cm 20 40 60 80 100 100 80 60 40 20 a)Student E made a mistake in the experiment since for a loss in mass of 88mg according to a line passing through the other points, volume should have been about 68cm3 instead of 76 cm3 b) The line of best fit passes through points (0,0) and (80, 60). The slope is m = 60/80 = 0.75 the equation of line is y = 0.75x c)Let y = 1000 d)1000= 0.75x, x = 1333 1/3 The mass of 1000cm3 of the gas is about 1333mg e)The density of unknown gas is 1333mg = 1333mg = 1333mg = 1333 mg/liter 1000cm3 1000militers 1 liter Since oxygen has a density of 1330mg/liter, oxygen is the most likely gas Hydrogen 8mg/liter Nitrogen 1160 mg/liter Oxygen 1330 mg/liter Carbon dioxide 1830 mg/liter

27. Using Graphing Calculator for Linear Regression Pg 59 Step 1 Press Stat, choose 1 press Enter Step 3 Stat, right arrow go to 4 And enter Step 4Press Vars 5,Right,Right, Enter Step 5 Press 2 nd, Stat Plot and enter Step 2 Enter Y = 1.95x – 7.86 Step 6 the graph ( Pg 60)

28. 1.6 Additional Properties of Lines ( pg 68) Horizontal Line y = 4 Horizontal line 4 X axis Y axis - 5 5 ( - 1, 4) (2, 4) y = k( const) Horizontal Line Slope = 0

29. Vertical Line X axis Y axis - 5 5 5 -5 x = 3 (3, 1) (3, 3) x = k (const) Vertical lineSlope is undefined

30. Perpendicular Lines Two lines with slopes m 1 and m 2 are perpendicular if and only if m 2 = - 1/m 1 5-5 5 y = 2/3 x -2 y = -3/2 x + 3 3 - 2 X axis Y axis m 1 = 2/3 m 2 = -3/2= - 1/ m 1 m 1 m2

31. Parallel Lines -5 5 5 y = 2/3 x + 2 y = 2/3 x -2 Two lines with slopes m 1 and m 2 are parallel if and only if m 1 = m 2 X axis Y axis

32. Show that the triangle with vertices A(0, 8), B(6, 2) and C(-4, 4) is a right triangle (-4, 4) (6, 2) (0, 8) A B C Slope of AB= m 1 = 2 – 8 = - 6 = - 1 6 - 0 6 Slope of AC =m 2 = 4 – 8 = - 4 = 1 - 1 = -1 = 1 = m 2 - 4 – 0 - 4 m 1 -1 AB is Perpendicular to AC X axis Y axis

33. Example 33 Pg 77 A) Sketch the triangle with vertices A(-6, -3), B ( - 6, 3), and C(4, 5). B) Find the slope of the side AC. C) Find the slope of the side of the altitude from point B to side AC D) Find an equation for the line that includes the altitude from point B to side AC A ( - 6, - 3) C ( 4, 5) B ( - 6, 3 ) 6 -6 -6 6 B ) Slope of side AC = - 3 – 5 = - 8 = 4 - 6 – 4 - 10 5 C) Slope of the altitude from Pt. B to side AC is perpendicular to AC and therefore – 5 4 D) y – 3 = -5/4 ( x + 6), y = 3/2 x + 15/2