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Extra Notes on Slides. Pre Calculus with Modelling and Visualization( 3 E) By Gary Rockswold. Ch 1.1 Numbers, Data, and Problem Solving. Important Definition Examples Natural Numbers – referred as counting numbers N = { 1,2,3,4,…. }

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  1. Extra Notes onSlides Pre Calculus with Modelling and Visualization( 3 E) By Gary Rockswold

  2. Ch 1.1 Numbers, Data, and Problem Solving Important DefinitionExamples Natural Numbers – referred as counting numbers N = { 1,2,3,4,…. } Integers - Includes the natural numbers, their opposites, and 0 ….., -2, -1, 0, 1, 2,….. Rational number -- Any number that can be expressed as the ratio of two integers p/q, where q = 0 , all repeating and all terminating decimals 2/1, 1/3, -1/4, 22/7, 1.2, 0.4369, – 0.25 = ¼, 0.33 = 1/3, and all fractions Irrational numbers Can be written as non-repeating, non-terminating decimals; cannot be rational numbers; a square root of A positive integer that is not an integer. Real Numbers – Represented by standard or decimal numbers, includes the rational numbers and irrational numbers Scientific notation A number in the form c x 10 n , where 1 < c < 10 and n is an integer, Used to represent numbers that are large or small in absolute value 0.789 = 7.89 x 10 -1 854,000 = 8.54 x 10 5 Percent change - If a quantity changes from c1 to c2, then the percent change equals c2 – c1 x 100 If the price of a gallon of gasoline changes from $1.00 to $ 1.40, then c1 the percent change is 1.40 – 1 x 100 1

  3. Ch 1.2 – Visualization of Data Mean or average – To find the mean or average of n numbers, divide their sum by n. The mean of the four numbers -2, 3, 4,5, 6,8 is -2 + 3 + 4 + 5 + 6 + 8 = 4 6 Range - The range of a list of data is the difference between the maximum and the minimum of the numbers or values. The range of the data 3, -6, 2, 8, 10 is 10 – (-6) = 16 (Maximum – Minimum = Range) Median – The median of a sorted list of numbers equals the value that is located in the middle of the list. Half the data is greater than or equal to the median, and half the data is less than or equal to the median. The median of 1,3,5,9, the average of the two middle values 3, 5. Therefore 3+5 = 4 is the median 2

  4. Ch 1.2 Relation is a set of ordered pairs. If the ordered pairs in a relation by (x, y). Then the set of all x-values is called the domain( D) of the relation and the set of all y-values is called the range (R). If the relation S = {(0, 3), ( 1,2), (1, 6), (2, 7), (3, 5) } The relation has domain D = { 0,1, 2, 3 } R = { 2, 3, 5, 6, 7 } Graphing a relation (2, 7) (1, 6) (3, 5) Scatter Plot ( 1, 2) Connecting the data points with straight-line segments called a line graph

  5. Distance Formula- The distance between (x1, y1) and (x2, y2) is d= (x2 = x1) 2+ (y2 – y1)2 Example - The distance between (1, -3) and (1, 4) d = = 7 Midpoint Formula – The midpoint of the line segment connecting ( x1, y1) and (x2, y2) is M= (x1 + x2, y1 + y2). 2 2 The mid point of the line segment connecting ( 3,4) and ( 1, -2) is M = ( 3 + 1, 4 – 2) =( 2, 1) 2 2 2 2 2 =

  6. Determine whether the triangle is isosceles ? (7, 1) 5 5 (3, 4) (0,0) Isosceles triangle has at least two equal sides The vertices of the given triangle are (0,0), (7,1), (3,4) d = ( 3- 0)2 + (4 – 0)2 = 3 2 + 4 2 = 9 + 16 = 25 = 5 ( Using distance formula ) d= ( 7 – 3) 2 + ( 1 – 4)2 = 4 2 + (-3) 2 = 16 + 9 = 25 = 5 The side between ( 0, 0) and (7, 1)and the side between (3, 4) and (7, 1) have equal length, so the triangle is isosceles.

  7. Midpoint Formula ( pg 27, no 46 ex 1.2) The population of US was 151 million in 1950 179 million in 1960 What will be the population in 1970 ? Solution Let y represent the population in 1970. The data point ( 1960, 179) as the midpoint between the data point (1950, 151) and (1970, y). The increase between 1950 and 1960 was 28 million. Therefore, the estimated population in 1970 is 179 + 28 = 207 million. The midpoint formula is M = ( 1950 + 1970 , 151 + y ) = (160, 179 ) 2 2 M = 151 + y = 179 or y = 207 million 2 207 million population in 1970

  8. Data involving two variables Pg 27, No 62 ( Ex 1.2 ) { (1, 1), (3, 0), (-5, -5), (8, -2), (0, 3) } • The Domain is D = {1,3, -5, 8, 0} and the range is R = {1, 0, -5, -2, 3} • The minimum x-value is -5, and the maximum x-value is 8. The minimum y-value is -5, and the maximum y-value is 3. • The axes must include -5< x < 8 and -5 < y < 3 4 2 -2 -4 (0, 3) (1, 1) (3, 0) -8 -6 -4 -2 0 2 4 6 8 (8, -2) (-5, -5)

  9. Function Notation The notation y = f(x) is called function notation. The input is x, the output is y, and the name of the function is f. Name y = f(x) Output Input variable y is called independent variable and variable x is called dependent variable A function computes exactly one output for each valid input. Expression f(16) = 4 is read “f of 16 equals 4 and indicates that f outputs 4 when the input is 16

  10. Representations of Functions Verbal representation (words) – Words describe precisely what is Computed- Example- f(x) = x + 3, 3 is added with Input x to get output Numerical representation (Table of values ) – A numerical representation is a table of values that lists input-output pairs for a Function- f(x) = 2x Diagrammatic Representation (Diagram) – Functions are sometimes represented using diagrams Symbolic Representation ( formula)- Mathematical formula – The function is defined by f(x) = x Graphical Representation (Graph) – Graph of ordered pairs (x, y) that satisfy y = f(x) A graph of f(x) = 2x Each point on the graph satisfy y = 2x Vertical line Test - If every vertical line intersects a graph at no more than one point, then the graph represents a function 1 2 1 2 3 5 7 9 5 6 7

  11. Use the graph of the function f to estimate its domain and range. Evaluate f(0)Ex 1.3 43 45 48 3 1 -1 -3 2 1 -1 -2 -2 -1 1 2 3 4 - 3 -1 1 3 -2 -1 1 2 Domain (D) = {x/ -3 < x < 3} Range (R) = { y/ 0< y < 3} f(0) = 3 Domain D = { x/ -2 < x < 4 Range R = { y / -2 < y < 2 }; f(0) = -2 Domain D = all real numbers Range R = all real numbers; f(0) = 0

  12. No 74 ( Ex 1.3)In 2004 the average cost of driving a new car was about 50 cents per mile. Give symbolic, graphical, and numerical representations that compute the cost in dollars of driving x miles. For the numerical representations that compute the cost in dollars of driving x miles. For the numerical representation use a table with x = 1, 2, 3, 4, 5, 6 0 20 40 60 80 100 Symbolic expression : f(x) = 0.50x Numerical 0 20 40 60 80 100 Graphical Cost of driving is $0.50 per mile

  13. ApplicationsNo. 105 ( Ex 1.3)When the relative humidity is 100%, air cools 5.80 F for every 1-mile increase in altitude. Give verbal, symbolic, graphical, and numerical representations of a function f that computes this change in temperature for an increase of x miles. Let the domain of f be 0< x < 3Verbal : Multiply the input x by – 5.8 to obtain the change in temperature Numerical :Table Y1 = - 5.8X Symbolic : f(x) = - 5.8 X Graphical : Y 1 = - 5.8X

  14. Types of Functions and their rates of Change Constant Function - A function f represented by f(x) = b, where b is constant ( fixed number ), is a constant function Linear Function – A function f represented by f(x) = ax + b, where a and b are constants, is a linear function Slope – The slope m of the line passing through the points (x1, y1 ) and (x2, y2 ) is y = y2 – y1 where x1 = x2 x x2 – x1 Example – a line passing through ( 1, 3) nand (2, 7) has slope m = 7-3 = 4 2 - 1 This slope indicates that the line rises 4 units for each unit increase in x m = 2 > 0 m = - ½ < 0 m = 0 m is undefined

  15. Non-Linear Functions If a function is not linear, then the function is called a nonlinear function. The graph of a nonlinear function is not a straight line f(x) = Square Root Function f(x) = x2 f(x) = x Absolute Value Function f(x) = x3 Cube Function

  16. Average Rate of Change Let (x1, y1) and (x2, y2 ) be distance points on the graph of a function f. The average rate of change of f from x1 to x2 ) is y2 – y1 x2 – x1 That is, the average rate of change from x1 to x2 equals the slope of the line passing through ( x1, y1) and (x2, y2) Difference Quotient The difference quotient of a function f is an expression of the form f(x + h) – f(x) where h = 0 h

  17. CH 2 Linear Functions and Equations Modeling with a linear function To model a quantity that is changing at a constant rate with a linear function f, the following may be used f(x) = (constant rate of change)x + (initial amount) Note : The constant rate of change corresponds to the slope of the graph of f and the initial amount corresponds to the y-intercept

  18. Correlation Coefficient r, (-1 < r < 1) Value of r Comments Sample Scatterplot r = 1 There is an exact linear fit. The line passes through all data points and has a positive slope r = -1 There is an exactlinear fit. The line passes through all data points and has a negative slope 0<r<1 There is a positive correlation. As the x-values increase, so do the y-values. The fit is not exact -1 < r < 0 There is a negative correlation. As the x-values increase, the y- values decrease. The fit is not exact r = 0 There is no correlation. The data has no tendency toward being linear. A regression line predicts poorly

  19. LINEAR REGRESSION • Find the line of least –squares fit for the data points ( 1, 1), (2, 3), and (3, 4). What is the correlation coefficient ? Plot the data and graph • Use Graphing Calculator • Hit STAT and EDIT Enter points (1, 1) , (2, 3), (3, 4) in the table Choose STAT and CALCLINEAR REGRESSION and ENTER Hit Y and ENTER a= 1.5 and b = - 1/3

  20. Hit STAT PLOTON will blink, Clear OFF (If it is black by using side arrows and enter) andchoose SCATTER PLOT and MARK using down arrow keys Hit WINDOW and Hit 2nd and Table enter points [ 0, 5, 1] by [0, 5, 1] Hit GRAPH

  21. Linear Regression Example 2Cellular Phones – One of the early problems with cellular phones was the delay involved with placing a call when the system was busy. One study analysed this delay. The table shows that as the number of calls increased by P percent, the average delay time D to put through a call also increased. P (%) 0 20 40 60 80 100 D( minutes) 1 1.6 2.4 3.2 3.8 4.4 Let P correspond to x-values and D to y-values. Find the least-squares regression line that models these data. Plot the data and the regression line Estimates the delay for a 50% increase in the number of calls Solution Use Graphing Calculator Hit STAT and EDIT Enter points (0, 1) , (20, 1.6), (40, 2.4), (60, 3.2), (80, 3.8), (100, 4.4) in the table

  22. Hit Y and enter a= 0.0349 and b= 0.9905 Choose STAT and CALC and LINEAR REGRESSION and enter Hit STAT PLOTON will blink, Clear OFF (If it is black by using s arrows and enter) andchoose SCATTER PLOT and MARK using down arrow keys Hit GRAPH Hit WINDOW and enter points [ -10, 110, 10 ] by [ 0, 5, 1]

  23. Point- Slope Form- the line with slope m passing through the point (x1, y1) has an equation y = m(x – x1) + y1 or y – y1 = m(x – x1) Slope- intercept form- The line with slope m and y-intercept b is given by y = mx + b, the slope- intercept form of the equation of a line. Finding intercepts – To find any x-intercepts, let y = 0 in the equation and solve for x To find any y-intercepts, let x = 0 in the equation and solve for y Equations of Horizontal and vertical lines – An equation of the horizontal Line with y-intercept b is y = b. An equation of the vertical line with x-intercept k is x = k Perpendicular lines- Two lines with nonzero slopes m1 and m2 are perpendicular if and only if their slopes have product – 1, that is m1m2 = -1

  24. Interpolation –Estimates that are between two or more known data values Extrapolation – Estimates values that are not between two known data values The table lists data that are exactly linear • Find the slope-intercept form of the line that passes through these data points. • Predict y when x = -2.7 and 6.3. Decide if these calculations involve interpolation or extrapolation No 54 ( Ex 2.2) Since the point (0, 6.8) is on the graph, the y-intercept is 6.8. The data is exactly linear, so one can use any two points to determine the slope. Using the points (0, 6.8) and (1, 5.1), m = 5.1- 6.8 = -1.7 ( slope formula , m = y2 – y1 1 – 0 x2 – x1 Slope intercept form of the line is y = - 1.7x + 6.8 ( y = mx+ b) b) When x = - 2.7, y = - 1.7(-2.7) + 6.8 = 11.39 This calculation involves extrapolation When x = 6.3, y = -1.7(6.3) + 6.8 = -3.91 This calculation involves extrapolation Y-intercept

  25. 2.3 – Linear EquationsLinear equation in one variable – A linear equation in one variable is an equation that can be written in the form ax + b = 0where a and b are real numbers with a= 0If the equation is not linear then the equation is nonlinear equationx – 4 = 0, 2x – 3 = 0, x – 4 + 3(x – 1) = 0 These are examples of linear equationsProperties of EqualityAddition Property of equalityIf a, b, and c are real numbers, then a = b is equivalent to a + c = b + cMultiplication Property of EqualityIf a, b, c are real numbers with c = 0 then a =b is equivalent to ac = bc

  26. Intersection of Graphs Method The intersection of graphs method can be used to solve the equation graphically. To implement this procedure, follow these steps Step 1 : Set y1 equal to the left side of the equation, and set y2 equal to the right side of the equation Step 2 : Graph y1 and y2 Step 3 : Locate any points of intersection. The x- coordinates of these points correspond to solutions to the equation

  27. Intermediate value Property Let (x1, y1) and (x2, y2) with y1= y2 and x1< x2 be two points on the graph of a continuous function f. Then , on the interval x1 < x < x2, f assumes every value between y1 and y2 at least once

  28. Solving Application problems Step 1 : Read the problem and make sure you understand it. Assign a variable to what you are being asked to find. If necessary, write other quantities in terms of this variable Step 2 : Write an equation that relates the quantities described in the problem. You may asked to sketch a diagram refer to known formulas Step 3 : Solve the equation and determine the solution Step 4 : Look back and check your solution. Does it seem reasonable ?

  29. Use the intersection of graphs method to solve the equation by hand. Check your answer ( Ex 2.3)51 ) x + 4 = 1 – 2x 52) 2x = 3x -1 y1 = x + 4, Y2 = 1 – 2x Let y1 = 2x y2 = 3x - 1 -4 -3 -2 -1 1 2 3 4 3 2 1 -1 -2 -3 3 1 -1 -3 ( 1, 2) -4 -3 -2 -1 1 2 3 4 The line intersects at x = - 1 The line intersects at x = 1

  30. Solve the linear equation with the intersection –of-graphs method. Approximate the solution to the nearest thousandth whenever appropriate ( Ex 2.3)No 58 No 60 No 638 – 2x = 1.6 = 4x – 6 6 – x = 2x – 3 7 3 Enter Y1= 8-2x Y2 = 1.6 Enter Y1= Y2 = 4x - 6 EnterY1 = (6 – x) / 7 Y2 = (2x – 3) /3 Hit Graph and calc, you will get graph and point of intersection

  31. Applications No 89 ( Ex 2.3 ). Conical Water Tank A water tank in the shape of an inverted cone has a height of 11 feet and a radius of 3.5 feet, as illustrated below, If the volume of the cone is V = 1/3 r2 h, Find the volume of the water in the tank when the water is 7 feet deep. 3.5 feet 11 feet Use similar triangles to find the radius of the cone when the water is 7 feet deep; r = 7 , ( Multiply by 3.5 ) , r = 7 (3.5) , r = 2.23 ft 3.5 11 11 V = 1 r2 h , 100 = 1 (3) 2 . h, 100 = 3 h 100 = h, h = 10.6 ft 3 3 3

  32. No 96 ( Ex 2.3)Perimeter Find the length of the longest side of the rectangle if its perimeter is 25 feet 2x 5x - 1 Perimeter P = 2w + 2L = 25 ( given) 2(2x) + 2(5x – 1) = 25 4x + 10x – 2 = 25 14x = 25 + 2 ( Add 2) 14x = 27 x = 27 ft = 1.93 ft 14 5x – 1 = 5( 27/4) – 1 ( substitute x value ) = 121/14 = 8.6 ft

  33. 2.4 Linear Inequalities Linear Inequality in one variable A linear inequality in one variable is an inequality that can be written in the form ax +b > 0 Where a = 0 ( The symbol > may be replaced by > , <, or < Examples of linear inequalities are 3x – 2 < 0, 2x + 5 > 6, x + 4 < 9 and 2x + 3 > -3x + 4

  34. Interval Notation Inequality Interval Notation Number line Graph - 2 < x < 2 ( - 2, 2) - 1 < x < 3 ( - 1, 3] - 3 < x < 2 [ - 3, 2 ] x < - 1 or x > 2 ( -  , - 1) U (2,  ) x > - 1 ( - 1,  ) x < 2 ( - , 2 ] ( ) -4 -3 -2 -1 0 1 2 3 4 ( ] -4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4 [ ] ( ) ) ] - < x <  -4 -3 -2 -1 0 1 2 3 4

  35. Properties of Inequalities Let a, b, c be real numbers.  • a < b and a+c < b+c are equivalent ( The same number may be added to or subtracted from both sides of an inequality.) Example To solve x, x – 7 < 5, x < 12( add 7 both sides ) • If c > 0, then a < b and ac < bc are equivalent. (Both sides of an inequality may be multiplied or divided by the same positive number) Example : To solve x 3x > 6, divide each by 3 we get, x > 2 • If c< 0, then a < b and ac > bc are equivalent. Each side of an inequality may be multiplied or divided by the same negative number provided the inequality symbol is reversed. Example : To solve x, -7x > 21, divide by -7 we get x < - 3 Note : Replacing < with < and > with > results in similar properties

  36. Compound Inequalities  Two inequalities connected by the word and or or Examples x < 3 or x > 4 x> - 3 and x < 4 The inequality x> 5 and x < 20 can be written as the three-part inequality 5< x < 20

  37. Ex 2.4 Solving Linear Inequalities Symbolically • -2(x – 10) + 1 > 0 -2x + 20 + 1 > 0 ( distribute -2 to remove parenthesis) -2x > -21 ( add -21 both sides) x < 21/2= 10.5 (Divide by -2 both sides, the inequality sign will change) ( - , 10.5) • 1 < 1 – 2t < 2 2 3 3 3 < 1 – 2t < 2 2 1 < - 2t < 1 2 - 1 > t >- 1 4 2 - 1 < t < 1 2 -4 ( -1 , - 1 ] 2 4

  38. Solving Linear Inequality Graphically ( Ex 2.4) 41 43 46 x + 2 > 2x 2x -2 > - 4 x + 4 -2 < 1-x < 2 3 3 Y1 = x + 2, Y2 = 2x Y1 = -2, Y2 = 1-x, Y3 = 2 Y1 = 2x -2 , Y2 = - 4 x + 4 3 3 [ -10, 10, 1] [ -10, 10, 1] [ -10, 10, 1] Y1 > Y2, when the line of Y1 is above the line of Y2, which by the graph is left of the intersection point ( 3, 0) implies { x / x> 3} Y1 > Y2, when the line of Y1 is above the line of Y2, which by the graph is left of the intersection point ( -1, 2) and (3, - 2) does not include { x/ -1 < x < 3} Y1 > Y2, when the line of Y1 is above the line of Y2, which by the graph is left of the intersection point ( 2, 4) implies { x/x< 2}

  39. Solve the compound linear inequality graphically. Write the solution set in interval notation, and approximate endpoints to the nearest tenth whenever appropriate Ex 2.4 • 66 3 < 5x – 17 < 15 0.2x < 2x -5 < 8 3 Y1 = 3, Y2 = 5x – 7 and Y3 = 15 Y1 = 0.2x, Y2 = 2x – 5, Y3 = 8 3 [ -5, 15, 5] by [ -5, 20, 5] Y1 = 3, Y2 = 5x – 17, Y3 = 15 Y1 = 0.2x , Y2 = (2x – 5)/3 , Y3 = 8

  40. No 70 ( Ex 2.4)Use the figure to solve each equation or inequalitya) f(x) = g(x)b) g(x) = h(x)f(x) < g(x) < h(x)d) g(x) > h(x) 700 600 500 400 300 200 100 y = g(x) y = h(x) y = f(x) 0 1 2 3 4 5 6 7 8 • f(x) = g(x) = 4b) g(x) = h(x) = 2c) f(x) < g(x) < h(x) • 2 <x < 4d) g(x) > h(x) • 0 < x < 2

  41. No 92 ( Ex 2.4) Motorcycles The number of Harley-Davidson motorcycles manufactured between 1985 and 1995 can be approximated by N(x) =6409(x – 1985) + 30,300, where x is the year • Did the demand for Harley- Davidson motorcycles increase or decrease over this time period ? Explain your reasoning. • Estimate the years when production was between 56,000 and 75,000 Solution • N(x) = 6409(x – 1985) + 30,300 Slope= m Slope of the graph of N is 6409. This means that Harley- Davidson has sold approximately 6409 more motorcycles each year. Demand has increased during this time period because the slope of N is positive • Graph Y1 = 6409(x – 1985) + 30,300, Y2 = 56,000, Y3 = 75,000 The points of intersection occur near (1989, 56,000) and (1992, 75,000). For 1989< x < 1992. Sales were between 56,000 and 75,000 motorcycles per year. That is, from 1989 to 1992 there were from 56,000 to 75,000 motorcycles sold per year Y3 Y2 Y1

  42. 2.5 Piece wise- defined function A function is piecewise defined if it has different formulas on different intervals of its domain. Many times the domain is restricted Examplesf(x) = 2x + 1 if -3 < x < 0 f(x) = 2 if - 5 < x < - 1 x - 1 if 0 < x < 3 x + 3 if -1 < x < 5 The domain of f { x/ - 3< x < 3} f(-2) = 2(-2) + 1= -3 The domain of f - 5 < x < 5 f(0) = 0 -1 = -1 f( -2) = 2 f(3)= 3 – 1 = 2 f(0) = 0 + 3 f is not continuous f(3)= 3 + 3 = 6 f is continuous 8 6 4 2 -2 8 6 4 2 -2 -6 -4 -2 2 4 6 -6 -4 -2 2 4 6

  43. Absolute value function f(x) = x , The output from the absolute value function is never negative, Examples f(3) = 3 = 3 f(6) = - 6 = 6 Absolute value equationsax + b = k with k > 0 is equivalent to ax + b = + k Example 3x – 2 = 4 3x – 2 = 4 or 3x – 2 = - 4 3x = 6 or 3x = -2 x = 2 or x = -2/3 If k < 0 , then the absolute value equation has no solution Absolute value inequalities Let the solutions to ax + b = k, k > 0 be s1 and s2 with s1 < s2 and the solution set to ax + b > k is x < s1 or x > s2 Examples To solve x – 5 < 3 or x – 5 >3 First solve x – 5 = -3 and x – 5 = 3 x = 2 x = 8 The solutions are 2 to 8 The solution set to x – 5 < 3 is 2 < x < 8 And the solution set to x – 5 > 3 is x < 2 or x > 8 V shaped Never go below the x-axis

  44. Alternative method for solving absolute value inequalities • ax + b < k is equivalent to –k < ax + b < k x - 1 < 4 is solved as follows 4 < x – 1 < 4 - 4 < x < 5 2. ax + b > k is equivalent to ax + b < -k or ax + b > k x – 1 > 3 is solved as follows x – 1 < -3 or x – 1 > 3 x < - 3 or x > 4

  45. Swimming Pool Levels The graph of y = f(x) shows the amount of water in thousands of gallons remaining in a swimming pool after x daysa) Estimate the initial and final amounts of water in the pool.b) When did the amount of water in the pool remain constantc) Approximate f(2) and f(4)d) At what rate was water being drained from the pool when 1 < x < 3 • Initial amount in the pool occurs when x = 0. f(0) = 50or 50,000 gallons. The final amount of water in the pool occurs when x = 5. Then f(5) = 30 or 30,000 gallons • The water level remained constant during the first day and the fourth day, when 0 < x < 1 or 3 < x < 4 • f(2) = 45 thousand and f(4) = 40 thousand • During the second and third days, the amount of water changed from 50,000 gallons to 40,000 gallons. This represents 10,000 gallons in 2 days or 5000 gallons per day were being pumped out of the pool 125 100 75 50 25 0 Gallons (thousands) 0 1 2 3 4 5 6 Time ( days)

  46. No 33 ( Ex 2.5) Graph y = f(x)Use the graph of y = f(x) to sketch a graph of the equation y = f(x)c) Determine the x intercept for the graph of the equation y = f(x) • A) The graph of Y1 = 2x • B) The graph of y = 2x is similar to the graph of y = 2x except that is reflected across x-axis whenever 2x < 0 The graph of Y1 = 2x • C) The x-intercept occurs when 2x = 0 or when x = 0. The x-intercept is located at (0,00

  47. Solve the absolute value equation ( Ex 2.5) • 51 - 3x – 2 = 5 4x – 5 + 3 = 2 Then – 3x – 2 = -5 4x – 5 = -1, has no solution -3x= - 5 + 2 ( add 2) since the absolute value of 3x = -3 any quantity is always greater x = 1 ( Divide by – 1) than or equal to 0 Or -3x – 2 = 5 -3x = 7 ( Add 2) x = - 7/3 ( Divide by – 3)

  48. No 61 ( Ex 2.5) Solve the equation graphically, numerically and symbolically a) Graph Y1 = abs(2x – 5) and Y2 = 10 b) Table Y1 = abs(2x -5) starting at -5, incrementing by 2.5 c) 2x – 5 = 10 2x – 5 = -10 Or 2x – 5 = 10 x = -2.5 or 7.5 From each method, the solution to 2x – 5 < 10 lies between – 2.5 and 7.5, -2.5 < x < 7.5

  49. No 95 ( Ex 2.5) Average temperatureAnalyze the temperature range in Boston MassachussettsThe inequality T – 50 < 22 describes the range of monthly average temperatures T in degree Fahrenheit.Solve the inequality graphically and symbolicallyb) The high and low monthly average temperatures satisfy the absolute value equation T – 50 = 22 • Graph Y1 = abs (T – 50) and Y2 = 22 • The V- shaped graph of y1 intersects the horizontal line at the points • T – 50 = -22 or T – 50 = 22 T = 28 or 72 • The average monthly temperature range is 28 0 F • B) The monthly average temperatures in Boston vary between a low of 280 F and a high of 720 F

  50. 3.1 Quadratic Function and Models Quadratic Function- Let a, b and c be real numbers with a= 0. A function represented by f(x) = ax2 + bx + c is a quadratic function Vertex Axis of Symmetry Axis of Symmetry

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