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Graphs, Linear Equations, and Functions• 3-1 The Rectangular Coordinate System• 3-2 The Slope of a Line• 3-3 Linear Equations in two variables• 3-5 Introduction to Functions 1
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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. 2
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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 y- axis does not belong to any quadrant (point E). The quadrants are numbered. 3
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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: Given: Equation Solution Solution X Y 3x - 4y =12 X Y 0 x=0 3(0) -4y = 12 y = -3 0 -3 0 y=0 3x - 4(0) =12 x=4 4 0 -12 y = -12 3x - 4(-12) = 12 x = -12 -4 -12 -12 x = -4 3(-4) -4y = 12 y = -6 -4 -6 4
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3-1 The Rectangular Coordinate System Graphing lines: Since an equation in two variables is satisfied with an infinite number of ordered pairs, it is common practice to “graph” an equation to give a picture of the solution.Note: An equation like 2x +3y = 6 is called a “first degree” equation because it has no term with a variable to a power greater than 1.The graph of any first degree equation in two variable is a straight line.• A linear equation in two variable can be written in the form: Ax + By = C, where A,B and C are real numbers (A and B both not zero). This form is called the “Standard Form”. 5
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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 6
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3-1 The Rectangular Coordinate SystemFinding 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) 7
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3-1 The Rectangular Coordinate SystemFinding Intercepts: In the equation of a line, let y = 0 to find the “x-intercept” and let x = 0 to find the “y-intercept”.Example 5: Find the intercepts and draw the graph of 4x –y = -3 x-intercept: Let y = 0 : 4x –0 = -3 4x = -3 x = -3/4 y-intercept: Let x = 0 : 4(0) – y = -3 -y = -3 y=3 3x-intercept is (- , 0) 4y-intercept is (0,3) 8
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3-1 The Rectangular Coordinate System Recognizing equations of vertical and horizontal lines: An equation with only the variable x will always intersect the x- axis and thus will be vertical. An equation with only the variable y will always intersect the y- axis and thus will be horizontal.• Example 6: A) Draw the graph of y = 3 B) Draw the graph of x + 2 = 0 x = -2A) B) 9
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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 10
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3-2 The Slope of a LineFinding the slope of a line given twopoints 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 = = = ( x 2 ≠ x1 ) run change in x x2 − x1Note: Be careful to subtract the y-values and the x-values in thesame order. Correct Incorrect y2 − y1 y1 − y2 y2 − y1 y1 − y2 or or x2 − x1 x1 − x2 x1 − x2 x2 − x1 11
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3-2 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 12
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3-2 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 slope- intercept form of a line. The slope is the coefficient of x and the other term is the y- intercept. 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 3 −4 y = −3 x + 12 ∴ The slope is 3 4 y = x −3 4 13
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3-2 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 = −3 x + 9 ∴ The slope is - 4 3 9 y =− x+ 4 4 14
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3-2 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 2slope = passing through (-1,4) 3Note: change in y is +2 15
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3-2 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 16
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3-2 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: For line 2: 5−2 3 10 − 7 3 m1 = = m2 = = 3 − (−1) 4 8−4 4 ∴ YES 17
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3-2 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: For line 2: 3x + 5 y = 6 5x - 3 y = 2 5 y = −3 x + 6 − 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 18
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3-2 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 19
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3-2 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 20
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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 y- intercept.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 y- intercept (0,-3)Since m = 2 and b = -3, y = 2x - 3 21
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3-3 Linear Equations in Two Variables Graphing a line using its slope and y- intercept:• Example 2: Graph the line using the slope and y- intercept: y = 3x - 6Since b = -6, one point on the line is (0,-6). 3Locate the point and use the slope (m = 1 ) to locate a second point.(0+1,-6+3)=(1,-3) 22
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3-3 Linear Equations in Two Variables Writing an equation of a line given its slope and a point on the line: The “Point- Slope” form of the equation of a line with slope m and passing through the point (x1,y1) is: y - y1 = m(x - x1) where m is the given slope and x1 and y1 are the respective values of the given point.• Example 3: Find an equation of a line with slope 2 and a given point (3,-4) y − y1 = m( x − x1 ) 5 2 y − (−4) = ( x − 3) 5 2 6 y+4= x− 5 5 2 6 20 2 26 y = x− − = x− 5 5 5 5 5 5 y = 2 x − 26 or in standard form 2 x − 5 y = 26 23
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3-3 Linear Equations in Two Variables Writing an equation of a line given two points on the line: The standard form for a line was defined as Ax + By = C.• Example 4: Find an equation of a line with passing through the points (-2,6) and (1,4). Write the answer in standard form. 4−6 −2 2 Step 1: Find the slope: m= = =− 1 − (−2) 3 3 Step 2: Use the point-slope y − y1 = m( x − x1 ) method: 2 y − 6 = − ( x − (−2)) 3 2 4 y−6 = − x− 3 3 2 4 18 2 14 y =− x− + =− x+ 3 3 3 3 3 3 y = −2 x + 14 or 2 x + 3 y = 14 24
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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 Step 2: Use the point-slope method of the given line y − y1 = m( x − x1 ) 2 2 x − 3 y = 10 y −3= 3 ( x − (−8)) −3 y = −2 x + 10 y −3= x+ 2 16 3 3 2 10 2 y= x− ∴ m= 2 16 9 2 y = x+ + = x+ 25 3 3 3 3 3 3 3 3 3 y = 2 x + 25 or -2 x + 3 y = 25 25
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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 Step 3: Use the point-slope method of the given line y − y1 = m( x − x1 ) 2 x − 3 y = 10 3 −3 y = −2 x + 10 y − 3 = − ( x − (−8)) 2 10 2 2 y = x− ∴ m= 3 3 3 3 y − 3 = − x − 12 Step 2: Take the negative reciprocal 2 of the slope found 3 y = − x−9 2 3 2 m= ∴ m1 = − 2 y = −3 x − 18 or 3x + 2 y = −18 3 2 26
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3-3 Linear Equations in Two VariablesForms of Linear Equations Equation Description When to Use Slope-Intercept Form Given an equation, the slope Y = mx + b slope is m and y-intercept can be easily y-intercept is (0,b) identified and used to graph This form is ideal to use when Point-Slope Formy - y1 = m(x-x 1) slope is m given the slope of a line and line passes through (x 1,y1) one point on the line or given two points on the line. Standard Form (A,B, and C are integers, A 0) ≥ Ax + By = C Slope is -(A/B) x-intercept is (C/A,0) X- and y-intercepts can be found y-intercept is (0,C/B) quickly Horizontal line y=b slope is 0 Graph intersects only the y y-intercept is (0,b) axis, is parallel to the x-axis Vertical line x=a slope is undefined Graph intersects only the x x-intercept is (a,0) axis, is parallel to the y-axis 27
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3-3 Linear Equations in Two VariablesWriting 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 28
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3-3 Linear Equations in Two VariablesWriting 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 1984 47 m= = 1988 51 1998 − 1980 18 1992 54 m = 1.167 1998 59 ∴ y = 1.167 x + 38 29
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3-5 Introduction to FunctionsDefining 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. 30
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3-5 Introduction to FunctionsDefining 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. 31
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3-5 Introduction to FunctionsDefining 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 y 0 4 2 8 4 12 6 16 32
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3-5 Introduction to FunctionsDomain 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} Range: {4,8,12,16} 0 4 2 8 4 12 6 16 33
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3-5 Introduction to FunctionsDomain 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? Cell Phone Users Domain: Subscribers {1994,1995,1996,1997,1998,1999} Year (thousands) 1994 24,134 Range: 1995 33,786 {24134,33786,44043,55312,69209,86047} 1996 44,043 This is a function 1997 55,312 1998 69,209 1999 86,047 34
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3-5 Introduction to FunctionsDomain 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] 35
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3-5 Introduction to FunctionsAgreement 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.• The function 1 y= 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. 2• The function y = 3 x − 2 is not defined for values < 3 36
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3-5 Introduction to FunctionsIdentifying functions defined by graphsand equations:Vertical Line Test:If every vertical line intersects the graph of a relationin no more than one point, the relation represents afunction . 37
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3-5 Introduction to FunctionsIdentifying functions defined by graphs andequations:Vertical Line Test:If every vertical line intersects the graph of a relation in nomore than one point, the relation represents a function. Function Not a Function 38
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3-5 Introduction to Functions Identifying functions defined by graphs and equations:• Example 5: Decide whether the equations shown define a function and give the domain 5 y = 3x − 2 y= y2 = x x −1 Function Function Not a Function Domain: Domain: Domain: 2 3 ,∞ ( −∞,1) U ( 1, ∞ ) [ 0, ∞ ) 39
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3-5 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 −3 x + 5• Example 6: f ( x) = 2 Find : f (−3) and f (r ) −3(−3) + 5 9 + 5 14 f (−3) = = = =7 2 2 2 −3r + 5 f (r ) = 2 40
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3-5 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) Find : f (1) and f (a ) (1) 2 − 3 −2 1 −4 y = − x + 3 f (1) = = =− 2 x2 − 3 4 4 2 y= 4 a2 − 3 f (a) = x2 − 3 4 f ( x) = 4 41
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3-5 Introduction to FunctionsIdentifying Linear Functions:A function that can be defined by f(x) = mx + b for realnumbers m and b is a “Linear Function”Note: 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.A function that can be defined by f(x) = b is called a “ConstantFunction” which has a graph that is a horizontal line.Note: The range of a constant function is {b}. 42
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