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Graphs, Linear Equations, and Functions • • • • • • • • 3-1 3-2 3-3 3-5 3.6 3.7 3.8 3.9 The Rectangular Coordinate System Graphs of Equations Lines Introduction to Functions Quadratic Functions Operations on Functions Inverse Functions Variation 1 The Rectangular Coordinate System Section 3.1 2 Objectives: • Identify parts of the rectangular coordinate system. • Graph linear equations using an x-y chart and using x and y intercepts. • Graph horizontal and vertical lines. • Use a graphing calculator to analyze data involving linear equations. • Apply the distance and midpoint formulas. 3 3-1 The Rectangular Coordinate System Plotting ordered pairs. An “ordered pair” of numbers is a pair of numbers written within parenthesis in which the order of the numbers is important. • Example 1: (3,1), (-5,6), (0,0) are ordered pairs. Note: The parenthesis used to represent an ordered pair are also used to represent an open interval. The context of the problem tells whether the symbols are ordered pairs or an open interval. Graphing an ordered pair requires the use of graph paper and the use of two perpendicular number lines that intersect at their 0 points. The common 0 point is called the “origin”. The horizontal number line is referred to as the “x-axis” or “abscissa” and the vertical line is referred to as the “y-axis” or “ordinate”. In an ordered pair, the first number refers to the position of the point on the x-axis, and the second number refers to the position of the point on the y-axis. 4 3-1 The Rectangular Coordinate System Plotting ordered pairs. The x-axis and the y-axis make up a “rectangular or Cartesian” coordinate system. • Points are graphed by moving the appropriate number of units in the x direction, than moving the appropriate number of units in the y direction. (point A has coordinates (3,1), the point was found by moving 3 units in the positive x direction, then 1 in the positive y direction) • The four regions of the graph are called quadrants. A point on the x-axis or yaxis does not belong to any quadrant (point E). The quadrants are numbered. 5 3-1 The Rectangular Coordinate System Finding ordered pairs that satisfy a given equation. To find ordered pairs that satisfy an equation, select any number for one of the variables, substitute into the equation that value, and solve for the other variable. • Example 2: For 3x – 4y = 12, complete the table shown: Solution Given: Equation Solution X 0 Y 0 -12 -4 x=0 3x - 4y =12 3(0) -4y = 12 y = -3 y=0 3x - 4(0) =12 x=4 y = -12 3x - 4(-12) = 12 x = -12 x = -4 3(-4) -4y = 12 y = -6 X 0 4 Y -3 0 -12 -12 -4 -6 6 Intercepts of a Line • The point where the line intersects the x-axis is (x, 0) and is called the x-intercept. To find the value of x, substitute 0 in for y and solve for x. • The point where the line intersects the y-axis is (0, y) and is called the y-intercept. To find the value of y, substitute 0 in for x and solve for y. 7 Graph the linear equation using x and y intercepts • Graph -2x + 4y = 8 8 Graph using either method • 2(x -1) = 6 - 8y 9 Equations of Horizontal and Vertical lines • If a and b are real numbers then: – The graph of x = a is a vertical line with x-intercept (a, 0). x = 3, x = -2, x = 7/9, x = -7.5 – The graph of y = b is a horizontal line with yintercept (0, b). y = 5, y = -1, y = 2/3, y = -3.4 10 Graph x = 2 Graph y = -3 11 The Distance Formula d ( x2 x1 ) ( y2 y1 ) 2 2 • Used to find the distance between any two points in a rectangular coordinate system. • The distance formula can be derived by plotting two points (x1, y1) and (x2, y2), then form a right triangle, and apply the Pythagorean theorem. 12 Find the distance between P(-2, -5) and Q(3, 7) 13 The Midpoint Formula x1 x2 y1 y2 , 2 2 •The point M that is half way or midway between points P(x1, x2) and Q(y1,, y2) is called the midpoint. •The midpoint is average of the x-coordinates and the average of the y-coordinates 14 Find the midpoint of the segment joining the points P(-7, -8) and Q(1, -4) 15 If the midpoint of the segment is M(2, -5) and one endpoint is P (6, 9), find the coordinates of the other endpoint Q. 16 Important Information • The graph of a linear equation is a straight line. • Know how to graph a linear equation using an x-y chart and using x and y intercepts. • Know that x = constant is a vertical line. • Know that y = constant is a horizontal line. • Know the distance formula and the midpoint formula. 17 3-1 The Rectangular Coordinate System Graphing lines: • Example 3: Draw the graph of 2x + 3y = 6 Step 1: Find a table of ordered pairs that satisfy the equation. Step 2: Plot the points on a rectangular coordinate system. Step 3: Draw the straight line that would pass through the points. Step 1 Step 2 Step 3 18 3-1 The Rectangular Coordinate System Finding Intercepts: In the equation of a line, let y = 0 to find the “x-intercept” and let x = 0 to find the “y-intercept”. Note: A linear equation with both x and y variables will have both x- and y-intercepts. • Example 4: Find the intercepts and draw the graph of 2x –y = 4 x-intercept: Let y = 0 : 2x –0 = 4 2x = 4 x=2 y-intercept: Let x = 0 : 2(0) – y = 4 -y = 4 y = -4 x-intercept is (2,0) y-intercept is (0,-4) 19 3-1 The Rectangular Coordinate System Recognizing equations of vertical and horizontal lines: An equation with only the variable x will always intersect the xaxis and thus will be vertical. An equation with only the variable y will always intersect the yaxis and thus will be horizontal. • Example 6: A) Draw the graph of y = 3 B) Draw the graph of x + 2 = 0 x = -2 A) B) 20 3-1 The Rectangular Coordinate System Graphing a line that passes through the origin: Some lines have both the x- and y-intercepts at the origin. Note: An equation of the form Ax + By = 0 will always pass through the origin. Find a multiple of the coefficients of x and y and use that value to find a second ordered pair that satisfies the equation. • Example 7: A) Graph x + 2y = 0 21 3.1 Homework Answers 15. d ( A, C ) d ( A, B) d ( B, C ) 2 2 25. d ( P, O) 5 2 ( x 0) 2 ( y 0) 2 5 2 ( 130 ) ( 98 ) 32 1 Area ( 32 )( 98 ) 28 2 2 2 17. d ( A, B) d ( B, C ) d (C, D) d ( D, A) Radius = 5, Center = (0,0) 29 d ( A, C ) 2 d ( A, B) 2 d ( B, C ) 2 d ( A, C ) 58 19. 3 x 5 2 8 y 10 2 27. d ( P, O ) 6 (0 5) 2 ( y 3) 2 36 25 y 2 6 y 9 y 3 11 (0,3 11) 29. 5 5x 2 y 3 and (0,3 11) (2a 1) 2 (a 3) 2 25 4a 2 4a 1 a 2 6a 9 0 (a 3)( a 1) 21. d ( A, C ) d ( B, C ) 145 23. d ( A, P) d ( B, P) ( x 4) 2 ( y 3) 2 ( x 6) 2 ( y 1) 2 Pick a pt. Q(0,y) on y-axis 38. Since y-coord. is – in the third quad. a = -1. (-2,-1) 1920 1940 2042 1878 , 2 2 1930,1960 3.2 Solving Polynomial Equations by Graphing 23 3-2 The Slope of a Line Finding the slope of a line given two points on the line: The slope of the line through two distinct points (x1, y1) and (x2, y2) is: rise change in y y2 y1 slope m run change in x x2 x1 ( x 2 x1 ) Note: Be careful to subtract the y-values and the x-values in the same order. Correct Incorrect y2 y1 y1 y2 or x2 x1 x1 x2 y2 y1 y1 y2 or x1 x2 x2 x1 24 Types of Equations • Quadratic - has the form ax2 + bx + c=0 • Highest exponent is two (this is the degree) • The most real solutions it has is two. 25 Types of Equations • Cubic - has the form ax3 + bx2 + cx + d = 0 • Highest exponent is three (this is the degree) • The most real solutions it has is three. 26 Types of Equations • Quartic - has the form ax4 + bx3 + cx2 + dx + e = 0 • Highest exponent is four (this is the degree) • The most real solutions it has is four. 27 Types of Equations • These keep on going up as the highest exponent increases. • You don’t need to know the names above quartic, but you do need to be able to give the degree. 28 29 The standard form of the equation of a circle with its center at the origin is x y r 2 2 2 r is the radius of the circle so if we take the square root of the right hand side, we'll know how big the radius is. Notice that both the x and y terms are squared. Linear equations don’t have either the x or y terms squared. Parabolas have only the x term was squared (or only the y term, but NOT both). 30 Let's look at the equation x y 9 2 2 This is r2 so r = 3 The center of the circle is at the origin and the radius is 3. Let's graph this circle. Count out 3 in all directions since that is the radius Center at (0, 0) -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 31 If the center of the circle is NOT at the origin then the equation for the standard form of a circle looks like this: x h y k 2 2 r The center of the circle is at (h, k). x 3 y 1 2 2 2 This is r2 so r = 4 16 Find the center and radius and graph this circle. The center of the circle is at (h, k) which is (3,1). The radius is 4 - - - - - - - 12345678 7 6 5 4 3 2 10 32 If you take the equation of a circle in standard form for example: 2 2 This is r2 so r = 2 x2 y4 4 (x - (-2)) Remember center is at (h, k) with (x - h) and (y - k) since the x is plus something and not minus, (x + 2) can be written as (x - (-2)) You can find the center and radius easily. The center is at (-2, 4) and the radius is 2. But what if it was not in standard form but multiplied out (FOILED) x 4 x 4 y 8 y 16 4 2 2 Moving everything to one side in descending order and combining like terms we'd have: x y 4 x 8 y 16 0 2 2 33 x y 4 x 8 y 16 0 2 2 If we'd have started with it like this, we'd have to complete the square on both the x's and y's to get in standard form. Group x terms and a place Group y terms and a place to complete the square to complete the square Move constant to the other side 2 4 16 16 ___ 4 ___ 16 x 4 x ____ y 8 y ____ 2 Complete the square Write factored and wahlah! back in standard form. x 2 y 4 2 2 4 34 Now let's work some examples: Find an equation of the circle with center at (0, 0) and radius 7. Let's sub in center and radius values in the standard form x 0h y k0 2 2 7r 2 x y 49 2 2 35 Find an equation of the circle with center at (0, 0) that passes through the point (-1, -4). Since the center is at (0, 0) we'll have x y r 2 2 2 The point (-1, -4) is on the circle so should work when we plug it in the equation: 1 4 2 2 r 2 1 16 17 Subbing this in for r2 we have: x y 17 2 2 36 Find an equation of the circle with center at (-2, 5) and radius 6 Subbing in the values in standard form we have: x -2h y k5 2 2 x 2 y 5 2 2 6r 2 36 37 Find an equation of the circle with center at (8, 2) and passes through the point (8, 0). Subbing in the center values in standard form we have: x 8h y k2 2 2 r 2 Since it passes through the point (8, 0) we can plug this point in for x and y to find r2. 8 8 0 2 2 2 r 2 x 8 y 2 2 2 4 4 38 Identify the center and radius and sketch the graph: 9 x 9 y 64 2 2 9 9 9 To get in standard form we don't want coefficients on the squared terms so let's divide everything by 9. 64 x y 9 2 2 Remember to square root this to get the radius. So the center is at (0, 0) and the radius is 8/3. - - - - - - - 12345678 7 6 5 4 3 2 10 39 Identify the center and radius and sketch the graph: x 4 y 3 2 2 25 Remember the center values end up being the opposite sign of what is with the x and y and the right hand side is the radius squared. So the center is at (-4, 3) and the radius is 5. - - - - - - - 01 234 56 78 765432 1 40 Find the center and radius of the circle: x y 6x 4 y 3 0 2 2 We have to complete the square on both the x's and y's to get in standard form. Group x terms and a place Group y terms and a place to complete the square to complete the square Move constant to the other side 2 9 4 3 ___ 9 ___ 4 x 6 x ____ y 4 y ____ 2 Write factored for standard form. x 3 y 2 2 2 16 So the center is at (-3, 2) and the radius is 4. 41 3.2 day 2 Homework Answers x 9 y2 x2 y2 9 33. 41. r = 3 43. (-4,4) 45. Midpt = (1,2) = center r 1 (diam.) 2 1 ( 136 ) 2 34 ( x 1) 2 ( y 2) 2 34 61. If distance from P to C is less than r, greater than r, or = to r 63. x – int. (0 for y) x2 4x 4 0 x2 y – int. (0 for x) y2 6 y 4 0 y 2 6 y 9 4 9 y 3 5 65.x 2 4 x 4 y 2 6 y 9 9 ( x 2) 2 ( y 3) 2 9 c(2,3) r 3 dP(2,6) d ( P, C ) 16 9 25 5 ( x 2) 2 ( y 3) 2 25 3-3 Lines The slope of a line Finding the slope of a line given two points on the line: • Example 1) Find the slope of the line through the points (2,-1) and (-5,3) rise y2 y1 3 (1) 4 4 slope m = run x2 x1 (5) 2 7 7 43 3-3 The Slope of a Line Finding the slope of a line given an equation of the line: The slope can be found by solving the equation such that y is solved for on the left side of the equal sign. This is called the slopeintercept form of a line. The slope is the coefficient of x and the other term is the yintercept. The slope-intercept form is y = mx + b • Example 2) Find the slope of the line given 3x – 4y = 12 3 x 4 y 12 4 y 3 x 12 3 y x3 4 3 The slope is 4 44 3-3 The Slope of a Line Finding the slope of a line given an equation of the line: • Example 3) Find the slope of the line given y + 3 = 0 y = 0x - 3 The slope is 0 • Example 4) Find the slope of the line given x + 6 = 0 Since it is not possible to solve for y, the slope is “Undefined” Note: Being undefined should not be described as “no slope” • Example 5) Find the slope of the line given 3x + 4y = 9 3x 4 y 9 3 4 y 3x 9 The slope is - 4 3 9 y x 4 4 45 3-3 The Slope of a Line Graph a line given its slope and a point on the line: Locate the first point, then use the slope to find a second point. Note: Graphing a line requires a minimum of two points. From the first point, move a positive or negative change in y as indicated by the value of the slope, then move a positive value of x. • Example 6) Graph the line given 2 slope = passing through (-1,4) 3 Note: change in y is +2 46 3-3 The Slope of a Line Graph a line given its slope and a point on the line: Locate the first point, then use the slope to find a second point. • Example 7) Graph the line given slope = -4 passing through (3,1) Note: A positive slope indicates the line moves up from L to R A negative slope indicates the line moves down from L to R 47 3-3 The Slope of a Line Using slope to determine whether two lines are parallel, perpendicular, or neither: Two non-vertical lines having the same slope are parallel. Two non-vertical lines whose slopes are negative reciprocals are perpendicular. • Example 8) Is the line through (-1,2) and (3,5) parallel to the line through (4,7) and (8,10)? For line 1: 52 3 m1 3 (1) 4 YES For line 2: 10 7 3 m2 84 4 48 3-3 The Slope of a Line Using slope to determine whether two lines are parallel, perpendicular, or neither: Two non-vertical lines having the same slope are parallel. Two non-vertical lines whose slopes are negative reciprocals are perpendicular. • Example 9) Are the lines 3x + 5y = 6 and 5x - 3y = 2 parallel, perpendicular, or neither? For line 1: 3x 5 y 6 5 y 3x 6 For line 2: 5x - 3 y 2 3 y 5 x 2 3 6 5 2 y x y x 5 5 3 3 3 5 is the negative reciprocal of 5 3 Perpendicular 49 3-3 The Slope of a Line Solving Problems involving average rate of change: The slope gives the average rate of change in y per unit change in x, where the value of y depends on x. • Example 10) The graph shown approximates the percent of US households owing multiple pc’s in the years 1997-2001. Find the average rate of change between years 2000 and 1997. Use the ordered pairs: (1997,10) and (2000,20.8) 20.8 10 10.8 % m 3.6 2000 1997 3 yr 50 3-3 The Slope of a Line Solving Problems involving average rate of change: The slope gives the average rate of change in y per unit change in x, where the value of y depends on x. • Example 11) In 1997, 36.4 % of high school students smoked. In 2001, 28.5 % smoked. Find the average rate of change in percent per year. Use the ordered pairs: (1997,36.4) and (2001,28.5) 28.5 36.4 7.9 % m 1.975 2001 1997 4 yr 51 3-3 Linear Equations in Two Variables Writing an equation of a line given its slope and y-intercept: The slope can be found by solving the equation such that y is solved for on the left side of the equal sign. This is called the slope-intercept form of a line. The slope is the coefficient of x and the other term is the yintercept. The slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. • Example 1: Find an equation of the line with slope 2 and yintercept (0,-3) Since m = 2 and b = -3, y = 2x - 3 52 3-3 Linear Equations in Two Variables Graphing a line using its slope and yintercept: • Example 2: Graph the line using the slope and yintercept: y = 3x - 6 Since b = -6, one point on the line is (0,-6). Locate the point and use the slope (m = 3 ) to locate a 1 second point. (0+1,-6+3)= (1,-3) 53 3-3 Linear Equations in Two Variables Finding equations of Parallel or Perpendicular lines: If parallel lines are required, the slopes are identical. If perpendicular lines are required, use slopes that are negative reciprocals of each other. • Example 5: Find an equation of a line passing through the point (-8,3) and parallel to 2x - 3y = 10. Step 1: Find the slope of the given line 2x – 3y = 10 -3y = -2x + 10 -3 -3 y = 2/3x – 10/3 m = 2/3 Step 2: Use slope-intercept form y = mx + b y = 2/3 x + b 3 = 2/3 (-8) + b 3 = -16/3 + b 25/3 = b y = 2/3x + 25/3 54 3-3 Linear Equations in Two Variables Finding equations of Parallel or Perpendicular lines: • Example 6: Find an equation of a line passing through the point (-8,3) and perpendicular to 2x - 3y = 10. Step 1: Find the slope of the given line 2 x 3 y 10 3 y 2 x 10 y 2 10 x 3 3 m 2 3 Step 2: Take the negative reciprocal Step 3: Use slope-intercept form y = mx + b y = -3/2x + b 3 = (-3/2)(-8) + b 3 = 12 + b -9 = b y = -3/2x - 9 of the slope found 2 3 m m1 3 2 55 3-3 Linear Equations in Two Variables Forms of Linear Equations Equation Description When to Use Y = mx + b Slope-Intercept Form slope is m y-intercept is (0,b) Given an equation, the slope and y-intercept can be easily identified and used to graph y - y1 = m(x-x 1) Point-Slope Form slope is m line passes through (x 1,y1) This form is ideal to use when given the slope of a line and one point on the line or given two points on the line. Standard Form (A,B, and C are integers, A 0) Slope is -(A/B) x-intercept is (C/A,0) y-intercept is (0,C/B) Horizontal line slope is 0 y-intercept is (0,b) X- and y-intercepts can be found quickly Ax + By = C y=b x=a Vertical line slope is undefined x-intercept is (a,0) Graph intersects only the y axis, is parallel to the x-axis Graph intersects only the x axis, is parallel to the y-axis 56 3-3 Linear Equations in Two Variables Writing an equation of a line that models real data: If the data changes at a fairly constant rate, the rate of change is the slope. An initial condition would be the y-intercept. • Example 7: Suppose there is a flat rate of $.20 plus a charge of $.10/minute to make a phone call. Write an equation that gives the cost y for a call of x minutes. Note: The initial condition is the flat rate of $.20 and the rate of change is $.10/minute. Solution: y = .10x + .20 57 3-3 Line of best fit Writing an equation of a line that models real data: If the data changes at a fairly constant rate, the rate of change is the slope. An initial condition would be the y-intercept. • Example 8: The percentage of mothers of children under 1 year old who participated in the US labor force is shown in the table. Find an equation that models the data. Year Percent Using (1980,38) and (1998,50) 1980 38 59 38 21 m 1998 1980 18 m 1.167 1984 1988 1992 1998 47 51 54 59 y 1.167 x 38 58 3.3 Homework Answers 11. A(-1,3) is 5 units to the left and 5 units downs from B(4,2). D will have the same relative position (-7-5,5-5) = (-12,0). 35. 1 5 5 5 1 midpt ( A, B) , b y 5 x 30 2 2 2 7 2 7 14 5 x 7 y 15 30 7 14 y 10 x 30 b slope ( A, B) 14 5 37. m=-1 thru origin and y = -x 47. (x-3)2+(y+2)2 = 49 53. a) 50 5 x x 14 b) 14 0 x 162 y 55.a) 20 m 3 b) 56. a) 5 (162) 58 14 20 w t 10 3 20 w t 10 50 c) 3 w 70 t 9 70 20 t 10 3 8125 m(1) 8250 p 125t 8250 b) P 5000 t 26 c) 8250 66 125 m 125 125t 8250 5000 3-4 Introduction to Functions Defining and Identifying Relations and Functions: If the value of the variable y depends upon the value of the variable x, then y is the dependent variable and x is the independent variable. • Example 1: The amount of a paycheck depends upon the number of hours worked. Then an ordered pair (5,40) would indicate that if you worked 5 hours, you would be paid $40. Then (x,y) would show x as the independent variable and y as the dependent variable. A relation is a “set of ordered pairs”. {(5,40), (10,80), (20,160), (40,320)} is a relation. 60 3-4 Introduction to Functions Defining and Identifying Relations and Functions: A function is a relation such that for each value of the independent variable, there is one and only one value of the dependent variable. Note: In a function, no two ordered pairs can have the same 1st component and different 2nd components. • Example 2: Determine whether the relation is a function {(-4,1), (-2,1), (-2,0)} Solution: Not a function, since the independent variable has more than one dependent value. 61 3-4 Introduction to Functions Defining and Identifying Relations and Functions: Since relations and functions are sets of ordered pairs, they can be represented as tables or graphs. It is common to describe the relation or function using a rule that explains the relationship between the independent and dependent variable. Note: The rule may be given in words or given as an equation. y = 2x + 4 where x is the independent variable and y is the dependent variable x 0 2 4 6 y 4 8 12 16 62 3-4 Introduction to Functions Domain and Range: In a relation: A) the set of all values of the independent variable (x) is the domain. B) the set of all values of the dependent variable (y) is the range. • Example 3: Give the domain and range of each relation. Is the relation a function? {(3,-1), (4,-2),(4,5), (6,8)} Domain: {3,4,6} Range: {-1,-2,5,8} Not a function • Example 4: Give the domain and range of each relation. Is the relation a function? x y Domain: {0,2,4,6} 0 4 Range: {4,8,12,16} 2 4 6 8 12 16 63 3-4 Introduction to Functions Domain and Range: In a relation: A) the set of all values of the independent variable (x) is the domain. B) the set of all values of the dependent variable (y) is the range. • Example 4: Give the domain and range of each relation. Is the relation a function? Domain: {1994,1995,1996,1997,1998,1999} Range: {24134,33786,44043,55312,69209,86047} This is a function Cell Phone Users Subscribers Year (thousands) 1994 24,134 1995 33,786 1996 44,043 1997 55,312 1998 69,209 1999 86,047 64 3-4 Introduction to Functions Domain and Range: In a relation: A) the set of all values of the independent variable (x) is the domain. B) the set of all values of the dependent variable (y) is the range. • Example 5: Give the domain and range of each relation. Domain: (-, ) Domain: [-4, 4] Range: (-, 4] Range: [-6, 6] 65 3-4 Introduction to Functions Agreement on Domain: Unless specified otherwise, the domain of a relation is assumed to be all real numbers that produce real numbers when substituted for the independent variable. 1 y • The function x has all real numbers except x = 0 Note: In general, the domain of a function defined by an algebraic expression is all real numbers except those numbers that lead to division by zero or an even root of a negative number. • The function y 3 x 2 is not defined for values < 2 3 66 3-4 Introduction to Functions Identifying functions defined by graphs and equations: Vertical Line Test: If every vertical line intersects the graph of a relation in no more than one point, the relation represents a function . 67 3-4 Introduction to Functions Identifying functions defined by graphs and equations: Vertical Line Test: If every vertical line intersects the graph of a relation in no more than one point, the relation represents a function. Function Not a Function 68 3-4 Introduction to Functions Identifying functions defined by graphs and equations: • Example 5: Decide whether the equations shown define a function and give the domain y 3x 2 y 5 x 1 y2 x Function Function Not a Function Domain: Domain: Domain: 2 3 , ,1 1, 0, 69 3-4 Introduction to Functions Using Function Notation: When a rule or equation is defined such that y is dependent on x, the “Function Notation” y = f(x) is used and is read as “y = f of x” where the letter f stands for function. Note: The symbol f(x) does not indicate that f is multiplied by x, but represents the y-value for the indicated x-value. If y = 9x -5, then f(x) = 9x -5 and f(2) = 9(2) -5 = 13 and f(0) = 9(0) -5 = -5 • Example 6: 3x 5 2 Find : f (3) and f (r ) 3(3) 5 9 5 14 f (3) 7 2 2 2 3r 5 f (r ) 2 f ( x) 70 3-4 Introduction to Functions Using Function Notation: When a rule or equation is defined such that y is dependent on x, the “Function Notation” y = f(x) is used and is read as “y = f of x” where the letter f stands for function. • Example 7: f(x) = 5x -1 find: f(m + 2) f(m + 2) = 5(m + 2) -1 = 5m + 10 - 1 f(m + 2) = 5m + 9 • Example 8: Rewrite the equation given and find f(1) and f(a) x2 4 y 3 Find : f (1) and f (a) 4 y x 3 2 x2 3 y 4 x2 3 f ( x) 4 Find : f (1) and f (a ) (1) 2 3 2 1 f (1) 4 4 2 a2 3 f (a) 4 71 3-4 Introduction to Functions Identifying Linear Functions: A function that can be defined by f(x) = mx + b for real numbers m and b is a “Linear Function” The domain of a linear function is (-, ). The range is (-, ). Note: Remember that m represents the slope of a line and (0,b) is the y-intercept. Note: A function that can be defined by f(x) = b is called a “Constant Function” which has a graph that is a horizontal line. Note: The range of a constant function is {b}. 72 3.4 Homework Answers 7. a) f (a) a 2 4 2 2 15. a) [-3,4] b) f (a) (a) 4 a 4 2 2 b) [-2,2] f ( a ) ( a 4 ) a 4 c) 2 2 c) f(1)=0 f ( a h ) a 2 ah h 4 d) d) x= -1, .5, 2 2 2 9. a) x x 3 a a 3 e) (-1, .5) or (2,4] 2 2 b) 1(a a 3) a a 3 31. D (, ) c) a 2 (a) 3 a 2 a 3 R (, ) 2 d) a h a h 3 Increas.(, ) e) (a 2 a 3) (h 2 h 3) 33. D (, ), f) (a 2ah h a h 3) (a a 3) R (,4) 2 2 2 h 4 11. a) 2 a 1 b) 4a 2 c) 4a d) 2a Increas.(,0] Decreas.[0, ) D [4, ) R [0, ) 35. Increas.[4, ) 37. D (, ) R {2} Const.(, ) D [6,6] 39. Dec.[6,0] R [6,0] Inc.[0,6] 73 Symmetric about the y axis FUNCTIONS Symmetric about the origin 3.5 Graphs of Functions 74 Even functions have y-axis Symmetry 8 7 6 5 4 3 2 1 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 -2 -3 -4 -5 -6 -7 So for an even function, for every point (x, y) on the graph, the point (-x, y) is also on the graph. 3.5 75 Odd functions have origin Symmetry 8 7 6 5 4 3 2 1 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 -2 -3 -4 -5 -6 -7 So for an odd function, for every point (x, y) on the graph, the point (-x, -y) is also on the graph. 3.5 76 x-axis Symmetry We wouldn’t talk about a function with x-axis symmetry because it wouldn’t BE a function. 8 7 6 5 4 3 2 1 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 -2 -3 -4 -5 -6 -7 3.5 77 A function is even if f( -x) = f(x) for every number x in the domain. So if you plug a –x into the function and you get the original function back again it is even. f x 5 x 2 x 1 4 2 Is this function even? YES f x 5( x) 2( x) 1 5x 2 x 1 4 2 4 2 f x 2 x x Is this function even? NO 3 3 f x 2( x) ( x) 2 x x 3.5 3 78 A function is odd if f( -x) = - f(x) for every number x in the domain. So if you plug a –x into the function and you get the negative of the function back again (all terms change signs) it is odd. f x 5 x 2 x 1 4 2 Is this function odd? NO f x 5( x) 2( x) 1 5x 2 x 1 4 2 4 2 f x 2 x x Is this function odd? YES 3 3 f x 2( x) ( x) 2 x x 3 79 If a function is not even or odd we just say neither (meaning neither even nor odd) Determine if the following functions are even, odd or neither. Not the original and all 3 terms didn’t change signs, so NEITHER. f x 5 x 1 f x 5 x 1 5 x 1 3 3 f x 3x x 2 4 2 Got f(x) back so EVEN. f x 3( x) ( x) 2 3x x 2 4 2 4 2 80 GREATEST INTEGER FUNCTION When greatest integer acts on a number, the value that represents the result is the greatest integer that is less than or equal to the given number. There are several descriptors in that expression. First of all you are looking only for an integer. Secondly, that integer must be less than or equal to the given number and finally, of all of the integers that satisfy the first two criteria, you want the greatest one. The brackets which indicate that this operation is to be performed is as shown: ‘[ ]’. Example: [1.97] = 1 Example: [-1.97] = -2 There are many integers less than 1.97; {1, 0, -1, -2, -3, -4, …} Of all of them, ‘1’ is the greatest. There are many integers less than -1.97; {-2, -3, -4, -5, -6, …} Of all of them, ‘-2’ is the greatest. 81 It may be helpful to visualize this function a little more clearly by 2 using a number line. 14 -6.31 -7 -6 5 -5 -4 -3 6.31 3 -2 Example: [6.31] = 6 -1 0 1 2 3 4 5 6 7 8 Example: [-6.31] = -7 When you use this function, the answer is the integer on the immediate left on the number line. There is one exception. When the function acts on a number that is itself an integer. The answer is itself. Example: [5] = 5 2 Example: 0 .6 0 3 Example: [-5] = -5 14 2 .8 3 Example: 5 If there is an operation inside the greatest integer brackets, it must be performed before applying the function. Example: [5.5–3.6] = [1.9] = 1 Example: [3.6–5.5] = [-1.9] = -2 Example: [5.5+3.6] = [9.1] = 9 Example: [5.53.6] = [19.8] = 19 82 2.01 1.99 1 2 3 Example: [5–2.99] = [2.01] = 2 Example: [5–3] = [2] = 2 Example: [5–3.01] = [1.99] = 1 The greatest integer function can be used to construct a Cartesian graph. The simplest of which is demonstrated below. f(x) = [x] To see what the graph looks like, it is necessary to determine some ordered pairs which can be determined with a table of values. x 0 1 2 3 -1 -2 f(x) = [x] f(0) = [0] = 0 f(1) = [1] = 1 f(2) = [2] = 2 f(3) = [3] = 3 f(-1) = [-1] = -1 If we only choose integer values for x then we will not really see the function manifest itself. To do this we need to choose noninteger values. f(-2) = [-2] = -2 83 x f(x) = [x] 0 f(0) = [0] = 0 0.5 f(0.5) = [0.5] = 0 0.7 f(0.7) = [0.7] = 0 0.8 f(0.8) = [0.8] = 0 0.9 f(0.9) = [0.9] = 0 1 f(1) = [1] = 1 1.5 f(1.5) = [1.5] = 1 1.6 f(1.6) = [1.6] = 1 1.7 f(1.7) = [1.7] = 1 1.8 f(1.8) = [1.8] = 1 1.9 f(1.9) = [1.9] = 1 2 f(2) = [2] = 2 -0.5 f(-0.5) =[-0.5]=-1 -0.9 f(-0.9) =[-0.9]=-1 -1 f(-1) = [-1] = -1 84 When all these points are strung together the graph looks something like this – a series of steps. For this reason it is sometimes called the ‘STEP FUNCTION’. Notice that the left of each step begins with a closed (inclusive) point but the right of each step ends with an open (excluding point) We can’t really state the last (most right) x-value on each step because there is always another to the right of the last one you may name. So instead we describe the first x-value that is NOT on a given step. Example: (1,0) 85 Rather than place a long series of points on the graph, a line segment can be drawn for each step as shown to the right. The graphs shown thus far have been magnified to make a point. However, these graphs are usually shown at a normal scale as you can see on the next slide. 86 f(x) = [x] In these 3 examples, parameter ‘a’ is changed. As a increases, the distance between the steps increases. a=1 f(x) = 2[x] a=2 a=3 f(x) = 3[x] 87 f(x) = -[x] a = -1 f(x) = -2[x] a = -2 When ‘a’ is negative, notice that the slope of the steps is changed. Downstairs instead of upstairs. But as ‘a’ changes from –1 to –2, the distance between steps increases. The further that ‘a’ is from 0, the greater the separation between steps. This can be described with a formula. Vertical distance between Steps = |a| 88 f(x) = [x] b=1 f(x) = [2x] b=2 1 f (x ) x 2 b 1 2 As ‘b’ is increased from 1 to 2, each step gets shorter. Then as it is decreased to 0.5, the steps get longer. 1 Length of Step b 89 3.5 Homework Answers 1. f ( x) 5( x) 2( x) f ( x) odd 3. f (x) 3(x) 2(x) 5 f ( x) even 39. a) shift 3 units left 5. f (x) 8(x) 3(x) f ( x) neither b) shift 3 units right c) shift up 3 units 7. f ( x) ( x) 4 f ( x) even d) shift down 3 units 9. f ( x) ( x) ( x) f ( x) odd 3 4 2 3 2 2 3 3 11. Shift down 2, up 1, 41. a) f is shifted left 9 units and up 1 b) f is reflected through x-axis up 3 c) f is reflected through x-axis and 13. Shift down 4, up 2, shifted left 7 units and down 1 up 4 53. Not a function (9) 15. Shift down 3, up 2 57. Graph 17. Shift left 2, right 3 67. Graph 25. P(0,5); y=f(x-2)-1 71. Since g(x) = f(x)+4, the graph of g (-2, 4) can be obtained by shifting 31. Shifted 2 units to the the graph of f up +4. right and 3 units up 91 Quadratic Equation – Equation in the form y=ax2 + bx + c. Parabola – The general shape of a quadratic equation. It is in the form of a “U” which may open upward or downward. Vertex – The maximum or minimum point of a parabola. Maximum – The highest point (vertex) of a parabola when it opens downward. Minimum – The lowest point (vertex) of a parabola when it opens upward. Axis of symmetry – The line passing through the vertex b about which the parabola is having the equation x symmetric. 2a 92 How does the sign of the coefficient of x2 affect the graph of a parabola? On your graphing calculator, do the following: 1. Press the Y= key. 2. Clear any existing equations by placing the cursor immediately after the = and pressing CLEAR. 3. Enter 2x2 after the Y1= by doing the following keystrokes. 4. Press GRAPH. 93 Repeat using the equation y = -2x2. When the coefficient of x2 is positive, the graph opens upward. When the coefficient of x2 is negative, the graph opens downward. 94 How does the value of a in the equation ax2 + bx + c affect the graph of the parabola? Clear the equations in the Y= screen of your calculator. Enter the equation x2 for Y1. Enter the equation 3 x2 for Y2. Choose a different type of line for Y2 so that you can tell the difference between them. Press GRAPH. 95 Clear the second equation in the Y= screen and now enter the equation y = (1/4)x2. Press the GRAPH key and compare the two graphs. 96 Summary for ax2 When a is positive, the parabola opens upward. When a is negative, the parabola opens downward. When a is larger than 1, the graph will be narrower than the graph of x2. When a is less than 1, the graph will be wider (broader) than the graph of x2. 97 How does the value of c affect the graph of a parabola when the equation is in the form ax2 + c? o In the Y= screen of the graphing calculator, enter x2 for Y1. o Enter x2 + 3 for Y2. o Press the GRAPH key. 98 Now predict what the graph of y = x2 – 5 will look like. Enter x2 for Y1 in the Y= screen. Enter x2 – 5 for Y2 Press GRAPH. 99 What happens to the graph of a parabola when the equation is in the form (x-h)2 or (x+h)2? Enter x2 for Y1 in the Y= screen. Enter (x-3)2 for Y2. Press GRAPH. 10 Clear the equation for Y2. Enter (x+4)2 for Y2. Press GRAPH. 10 The vertex of the graph of ax2 will be at the origin. The vertex of the graph of the parabola having the equation ax2 + c will move up on the y-axis by the amount c if c>0. The vertex of the graph of the parabola having the equation ax2 + c will move down on the y-axis by the absolute value of c if c<0. The vertex of the graph of the parabola in the form (x-h)2 will shift to the right by h units on the x-axis. The vertex of the graph of the parabola in the form (x+h)2 will shift to the left by h units on the x-axis. 10 Compare the graphs of the following quadratic equations to each other. Check your work with your graphing calculator. 1) x2, x2 – 7, (x +2)2 2) 2x2, x2 + 6, (1/3)(x-5)2 10 Problem 1 All three graphs have the same shape. The vertex of the graph of x2 – 7 will move down 7 on the y-axis. the vertex of the graph of (x+2)2 will move left two on the x-axis. 10 Problem 2 The graph of 2x2 will be the narrowest. The graph of (1/3)(x-2)2 will be the broadest. The vertex of x2 + 6 will be shifted up 6 units on the yaxis compared to the graph of 2x2. The vertex of (1/3)(x-2)2 will be shifted right two units on the x-axis compared to the graph of 2x2. 10 GRAPHS OF QUADRATIC FUNCTIONS When the standard form of a quadratic function f(x) = ax2 + bx + c is written in the form: a(x - h) 2 + k We can tell by horizontal and vertical shifting of the parabola where the vertex will be. The parabola will be shifted h units horizontally and k units vertically. 10 GRAPHS OF QUADRATIC FUNCTIONS Thus, a quadratic function written in the form a(x - h) 2 + k will have a vertex at the point (h,k). The value of “a” will determine whether the parabola opens up or down (positive or negative) and whether the parabola is narrow or wide. 10 GRAPHS OF QUADRATIC FUNCTIONS a(x - h) 2 + k Vertex (highest or lowest point): (h,k) If a > 0, then the parabola opens up If a < 0, then the parabola opens down 10 GRAPHS OF QUADRATIC FUNCTIONS Axis of Symmetry The vertical line about which the graph of a quadratic function is symmetric. x=h where h is the x-coordinate of the vertex. 10 GRAPHS OF QUADRATIC FUNCTIONS So, if we want to examine the characteristics of the graph of a quadratic function, our job is to transform the standard form f(x) = ax2 + bx + c into the form f(x) = a(x – h)2 + k 11 GRAPHS OF QUADRATIC FUNCTIONS This will require to process of completing the square. 11 GRAPHING QUADRATIC FUNCTIONS Graph the functions below by hand by determining whether its graph opens up or down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts, if any. Verify your results using a graphing calculator. f(x) = 2x2 - 3 g(x) = x2 - 6x - 1 h(x) = 3x2 + 6x k(x) = -2x2 + 6x + 2 11 EXAMPLE Determine without graphing whether the given quadratic function has a maximum or minimum value and then find the value. Verify by graphing. f(x) = 4x2 - 8x + 3 g(x) = -2x2 + 8x + 3 11 FINDING A QUADRATIC FUNCTION Determine the quadratic function whose vertex is (1,- 5) and whose y-intercept is -3. 11 3.6 Homework Answers 3 2 y 34 ( x 12 x 36) 11. 4 23. v(4,1) y y a( x 4) 1 2 25. v(2,4) pt (1,0) 3 ( x 6) 2 7 4 1 a 8 y a( x 2) 2 4 a 2 25 a ( 3 0 ) 2 27. 4 9 a3 0 a(0 3) 2 5 a 5 9 1 y ( x 4) 2 1 8 y 4 ( x 2) 2 4 9 y 3x 2 2 y 29. pt (0,0) 31. v(1,4), x (3,0) 0 a(3 1) 2 4 a 5 ( x 3) 2 5 9 1 4 y 1 ( x 1) 2 4 4 4 9 45. y ( x ) 3 27 2 60. a) up slightly, down slightly, although there is unexpected down in 1987 2 b)y .07( x 1984) 7.6 11 The sum f + g f g x f x gx This just says that to find the sum of two functions, add them together. You should simplify by finding like terms. f x 2 x 3 g x 4 x 1 2 3 f g 2x 3 4x 1 2 3 4x 2x 4 3 2 Combine like terms & put in descending order 11 The difference f - g f g x f x gx To find the difference between two functions, subtract the first from the second. CAUTION: Make sure you distribute the – to each term of the second function. You should simplify by combining like terms. f x 2 x 3 2 g x 4 x 1 3 f g 2x 3 4x 1 2 3 Distribute negative 2 x 3 4 x 1 4 x 2 x 2 2 3 3 2 11 The product f • g f g x f x g x To find the product of two functions, put parenthesis around them and multiply each term from the first function to each term of the second function. f x 2 x 3 g x 4 x 1 2 3 f g 2x 3 4x 1 2 3 8 x 2 x 12 x 3 5 2 3 FOIL Good idea to put in descending order but not required. 11 The quotient f /g f f x x g x g To find the quotient of two functions, put the first one over the second. f x 2 x 3 2 f 2x 3 3 g 4x 1 2 g x 4 x 1 3 Nothing more you could do here. (If you can reduce these you should). 12 So the first 4 operations on functions are pretty straight forward. The rules for the domain of functions would apply to these combinations of functions as well. The domain of the sum, difference or product would be the numbers x in the domains of both f and g. For the quotient, you would also need to exclude any numbers x that would make the resulting denominator 0. 12 COMPOSITION FUNCTIONS “SUBSTITUTING ONE FUNCTION INTO ANOTHER” 12 The Composition Function f g x f gx This is read “f composition g” and means to copy the f function down but where ever you see an x, substitute in the g function. f x 2 x 3 2 g x 4 x 1 3 f g 24 x 1 3 3 2 FOIL first and then distribute the 2 32 x 16 x 2 3 32 x 16 x 5 6 3 6 3 12 g f x g f x This is read “g composition f” and means to copy the g function down but where ever you see an x, substitute in the f function. f x 2 x 3 g x 4 x 1 2 3 g f 42 x 3 1 2 3 You could multiply this out but since it’s to the 3rd power we won’t 12 f f x f f x This is read “f composition f” and means to copy the f function down but where ever you see an x, substitute in the f function. (So sub the function into itself). f x 2 x 3 g x 4 x 1 2 3 f f 22 x 3 3 2 2 12 The DOMAIN of the Composition Function The domain of f composition g is the set of all numbers x in the domain of g such that g(x) is in the domain of f. f g 1 f x x 1 g x x 1 The domain of g is x 1 x 1 domain of f g is x : x 1 We also have to worry about any “illegals” in this composition function, specifically dividing by 0. This would mean that x 1 so the domain of the composition would be combining the two restrictions. 12 3.7 Homework Answers 1. f (3) g (3) 6 9 15 3. f (3) g (3) 6(9) 54 f ( x) g ( x) 3x 2 1 f (3) g (3) 6 9 3 15. f (3) 6 2 g (3) 9 3 f ( x) g ( x) 3 x 2 5. f ( x) g ( x) 2 x 4 3x 2 2 f ( x) g ( x) 2 x 5 9. 11. 13. f ( x) g ( x) x 5 f ( x) x 2 2 17. g ( x) 2 x 2 1 f ( x) g ( x) 0 f ( x) 1 g ( x) h( x ) 8 x 2 x 5 k ( x ) 4 x 6 x 9 h(2) 31 k (3) 45 2 2 h( x) 8x3 20 x k ( x) 128x3 20 x h(2) 24 k (3) 3396 19. h( x) 7 k ( x) 7 h(2) 7 k (3) 7 21. All R’s such that x>=-2 Domain: f ( f ( x)) (4 x 2) 1 g ( g ( x)) ( x 4 ) (x<=1) or (x>=2) h( x ) 6 x 9 k ( x) 6 x 8 35. f ( g ( x)) 2 x 2 1 g ( f ( x)) (4 x 2 4 x 1) h(2) 3 k (3) 10 x 3 2 h( x) 75 x 2 4 k ( x) 15 x 2 20 37. f(g(6))=5, g(f(6))=6, f(f(6))=6, and h(2) 304 k (3) 155 g(g(6))=5 12 3.8 One-to-One Functions; Inverse Function 12 A function f is one-to-one if for each x in the domain of f there is exactly one y in the range and no y in the range is the image of more than one x in the domain. A function is not one-to-one if two different elements in the domain correspond to the same element in the range. 12 x1 y1 x2 x3 y2 Domain Range x1 y1 x2 x3 y3 y3 Domain One-to-one function x1 x3 Domain Range NOT One-to-one function y1 y2 y3 Range Not a function 13 M: Mother Function is NOT one-one Joe Samantha Laura Anna Julie Ian Hilary Chelsea Barbara George Sue Humans Mothers 13 S: Social Security function IS one-one Joe 123456789 Samantha 223456789 Anna 333456789 Ian 433456789 Chelsea 533456789 George 633456789 Americans SSN 13 Is the function f below one – one? 1 10 2 11 3 12 4 13 5 14 6 15 7 16 13 Theorem Horizontal Line Test If horizontal lines intersect the graph of a function f in at most one point, then f is one-to-one. 13 Use the graph to determine whether 2 the function f ( x ) 2 x 5 x 1 is one-to-one. Not one-to-one. 13 Use the graph to determine whether the is one-to-one. function One-to-one. 13 The inverse of a one-one function is obtained by switching the role of x and y 13 S 1 The inverse of the social security function 123456789 Joe 223456789 Samantha 333456789 Anna 433456789 Ian 533456789 Chelsea 633456789 George SSN Americans 13 Let f denote a one-to-one function y = f(x). . The inverse of f, denoted by f -1 , is a function such that f -1(f( x )) = x for every x in the domain of f and f(f -1(x))=x for every x in the domain of f -1. 13 Theorem The graph of a function f and the 1 graph of its inverse f are symmetric with respect to the line y = x. 14 y=x f 6 4 f (0, 2) 1 2 (2, 0) 2 0 2 4 6 2 14 Finding the inverse of a 1-1 function Step1: Write the equation in the form y f (x) Step2: Interchange x and y. Step 3: Solve for y. Step 4: Write for y. f 1 ( x) 14 Find the inverse of Step1: 5 f ( x) x 3 5 y x3 Step2: Interchange x and y Step 3: Solve for y 5 x y 3 14 5 3x f ( x) x 1 14 14 14 3.9 Direct & Inverse Variation 14 Example 1. If W varies directly with F and when W = 24 , F = 6 . Find the value W when F = 10. W 4F Solution. W F W kF When W = 24, F = 6 24 k 6 When F = 10 W = ? W 4 10 W 40 k 6 24 k 4 14 Example 2. If g varies directly with the square of h and when g = 100 , h = 5 . Find the value h when g = 64. g 4h Solution. g h 2 g kh When g = 64 , h = ? 2 When g = 100 , h = 5 100 k 25 k 25 100 k4 2 64 4h 2 4h2 64 h 2 16 h 16 h = 4 or h = - 4 14 Example 3. If d varies inversely with w and when d = 3 , w = 9 . Find the value d when w = 3. Solution. 1 d w k d w 27 d w When w = 3 d = ? 27 d 3 When d = 3 , w = 9 k 3 9 d=9 k = 27 15 Example 4 . If r varies inversely with the square root of f and when r = 32 , f = 16. Find f when r = 32. Solution. r 1 f r k f When r = 32 , f= 16. k 16 32 32 k 4 k = 128 128 r f When r = 32 , f = ? 32 128 f 32 f 128 128 f 32 f 4 f 16 15 Example 5 . If t varies jointly with m and b and t = 80 when and b = 5. Find t when m = 5 and b = 8 . Solution. t mb m=2 t 8mb When m = 5 , b = 8 , t = ? t kmb t 8 5 8 When t = 80 , m = 2 and b = 5 t 320 80 k 2 5 10k 80 k 8 15 Example 6 . c varies directly with the square of m and inversely with w. c = 9 when m = 6 and w = 2 . Find c when m = 10 and w=4. Solution. m2 c w c km 2 w When c = 9 , m = 6 and w = 2 9 k 62 2 36k 9 2 1 K=½ m2 c 2w When m = 10 , w = 4 and c = ? 10 2 =12.5 c 2 4 15 Examination Questions. Example 1. The time,T minutes ,taken for a stadium to empty varies directly as the number of spectators , S, and inversely as the number of open exits, E. (a) Write down a relationship connecting T,S and E. It takes 12 minutes for a stadium to empty when there are 20 000 spectators and 20 open exits. (b) How long does it take the stadium to empty when there are 36 000 spectators and 24 open exits ? 15 Solution. 6S T 500 E (a) T S E K S T E K is the constant of variation. Now S = 36 000 and E = 24 . (b) 6 36000 T 500 24 T = 12 , S = 20 000 and E = 20 Substitute to find the value of K. 12 K 20000 1 20 Cross multiply. Substitute. T = 18 minutes 20 000 K = 20 x 12 240 6 K 20000 500 15 Example 2. The number of letters, N , which can be typed on a sheet of paper varies inversely as the square of the size, S , of the letters used. (a) Write down a relationship connecting N and S . (b) The size of the letters used is doubled. What effect does this have on the number of letters which can be typed on the sheet of paper ? Solution. (a) (b) Letter size = 2S 1 N 2 S K N (2 S )2 K N 2 S N K 4S 2 By doubling the size of letters the number of letters is quartered. 15 Example 3. A frictional force is necessary for a car to round a bend. The frictional force , F kilonewtons , varies directly as the square of the car’s speed , V metres per second, and inversely as the radius of the bend, R metres. (a) Write down a relationship between F, V and R. A frictional force of 20 kilonewtons is necessary for a car , travelling at a given speed , to round a bend. (b) Find the frictional force necessary for the same car , travelling at twice the given speed , to round the same bend. 15 Solution. (a) V2 F R KV 2 F R (b) Let the speed = 2V K ( 2V )2 F R 4 KV 2 F R By doubling the speed the frictional force F required to round the bend becomes 4 times greater. F = 4 x 20 = 80 K 4V 2 F R 15 3.9 Homework Answers 1. 3. u kv 12 k (30) 2 k 5 s rk t 5. k 14 13. (a) p kd (b) 118 k (2) k 59 (c) p (59)5 295 k (2) 7 4 kx2 k (25) k 3 25 3 27 7. k 27 z kx2 y 3 16 k (49) (8) 9. 2 k 49 11. y kx 16 k (4) z2 9 k 36 y k x z3 5 k (3) 8 k 40 3 kl R 2 d k (100) 15. (a) (b) 25 (.01) (c) 50 2 k 1 4000 9 17. (a) p k (b) 1.5 k (c) 3 2 l 2 k 3 2 4 3 15 Ch. 3 Quiz Review Answers Ch. 3 Test Packet Answers