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Chapter 2

2-1 Relations and Functions. A set of ordered pairs forms a relation.Example: {(2, 4) (0, 3) (4, -2) (-1, -8)}The domain is the set of all the first coordinates (x-coordinate) and the range is the set of all the second coordinates (y-coordinate).Example: domain: {2, 0, 4, -1} and range: {4, 3, -2, -8}Mapping shows how each member of the domain and range are paired.Example: 1943 -3210 7-5-2-67.

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Chapter 2

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    1. Chapter 2 Graphing Linear Relations and Functions

    3. 2-1 Relations and Functions (cont.) A function is a relation where an element from the domain is paired with only one element from the range. Example (from mapping example): The first is a function, but the second is not because the 1 is paired with both the 3 and the 0. If you can draw a vertical line everywhere through the graph of a relation and that line only intersects the graph at one point, then you have a function.

    4. 2-1 Relations and Functions (cont.) A discrete function consists of individual points that are not connected. When the domain of a function can be graphed with a smooth line or curve, then the function is called continuous.

    5. 2-1 Practice Find the domain and range of the following: {(3, 6) (-1, 5) (0, -2)} {(4, 1) (1, 0) (3, 1) (1, -2)} Are the following functions? If yes, are they discrete or continuous? {(2, 2) (3, 6) (-2, 0) (0, 5)} c. y = 8x2 + 4 {(9, 3) (8, -1) (9, 0) (9, 1) (0, -4)}

    6. 2-2 Linear Equations A linear equation is an equation whose graph is a straight line. The standard form of a linear equation is: Ax + By = C, where A, B, and C are all integers and A and B cannot both be 0. Linear functions have the form f(x) = mx + b, where m and b are real numbers. A constant function has a graph that is a straight, horizontal line. The equation has the form f(x) = b

    7. 2-2 Linear Equations (cont.) The point on the graph where the line crosses the y-axis is called the y-intercept. Example: find the y-intercept of 4x – 3y = 6 4(0) – 3y = 6 ? substitute 0 for x y = -2, so the graph crosses the y-axis at the point (0, -2) The point on the graph where the line crosses the x-axis is called the x-intercept. Example: find the x-intercept of 3x + 5y = 9 3x + 5(0) = 9 ? substitute 0 for y x = 3, so the graph crosses the x-axis at the point (3, 0)

    8. 2-2 Practice Determine if the following are linear equations. If so, write the equation in standard form and determine A, B, and C. 4x + 3y = 10 c. 5 – 3y = 8x x2 + y = 2 d. 1/x + 4y = -5 Find the x- and y-intercepts of the following: 4x – 3y = -12 ˝ y + 2 = ˝ x

    9. 2-3 Slope The slope of a line is the change in y over the change in x. If a line passes through the points (x1, y1) and (x2, y2), then the slope is given by m = y2 – y1 , where x1 ? x2. x2 – x1 In an equation with the from y = mx + b, m is the slope and b is the y-intercept. Two lines with the same slope are parallel. If the product of the slopes of two lines is -1, then the lines are perpendicular.

    10. 2-3 Practice Find the slope of the following: (-2, 4) (3, -6) d. (-1, 8) (14, 8) (3.5, -2) (0, -16) e. 12x + 3y – 6 = 0 y = 3x + b f. y = -7 Determine whether the following lines are perpendicular or parallel by finding the slope. (4, -2) (6, 0), (7, 3) (6, 2) y = 2x – 3, (6, 6) (4, 7)

    11. 2-4 Writing Linear Equations The form y = mx + b is called slope-intercept form, where m is the slope and b is the y-intercept. The point-slope form of the equation of a line is y – y1 = m(x – x1). Here (x1, y1) are the coordinates of any point found on that line.

    12. 2-4 Writing Linear Equations Example: Find the slope-intercept form of the equation passing through the point (-3, 5) with a slope of 2. y = mx + b 5 = (2)(-3) + b 5 = -6 + b b = 11 y = 2x + 11

    13. 2-4 Writing Linear Equations Example: Find the point-slope form of the equation of a line that passes through the points (1, -5) and (0, 4). m = y2 – y1 y – y1 = m(x – x1) x2 – x1 y – (-5) = (-1)(x – 1) m = 4 + 5 y + 5 = -x + 1 0 – 1 y = -x – 4 m = 9 -1 m = -1

    14. 2-4 Practice Find the slope-intercept form of the following: a line passing through the point (0, 5) with a slope of -7 a line passing through the points (-2, 4) and (3, 14) Find the point-slope form of the following: a line passing through the point (-2, 6) with a slope of 3 a line passing through the points (0, -9) and (-2, 1)

    15. 2-5 Modeling Real-World Data Using Scatter Plots Plotting points that do not form a straight line forms a scatter plot. The line that best represents the points is the best-fit line. A prediction equation uses points on the scatter plot to approximate through calculation the equation of the best-fit line.

    16. 2-5 Practice Plot the following data. Approximate the best-fit line by creating a prediction equation.

    17. 2-6 Special Functions Whenever a linear function has the form y = mx + b and b = 0 and m ? 0, it is called a direction variation. A constant function is a linear function in the form y = mx + b where m = 0. An identity function is a linear function in the form y = mx + b where m = 1 and b = 0.

    18. 2-6 Special Functions Step functions are functions depicted in graphs with open circles which mean that the particular point is not included. Example:

    19. 2-6 Special Functions A type of step function is the greatest integer function which is symbolized as [x] and means “the greatest integer not greater than x.” Examples: [8.2] = 8 [3.9] = 3 [5.0] = 5 [7.6] = 7 An absolute value function is the graph of the function that represents an absolute value. Examples: |-4| = 4 |-9| = 9

    20. 2-6 Practice Identify each of the following as constant, identity, direct variation, absolute value, or greatest integer function h(x) = [x – 6] e. f(x) = 3|-x + 1| f(x) = -˝ x f. g(x) = x g(x) = |2x| g. h(x) = [2 + 5x] h(x) = 7 h. f(x) = 9x Graph the equation y = |x – 6|

    21. 2-6 Answers Answers: 1)a) greatest integer function b) direct variation c) absolute value d) constant e) absolute value f) identity g) greatest integer function h) direct variation 2)

    22. 2-7 Linear Inequalities Example: Graph 2y – 8x = 4 Graph the “equals” part of the equation. 2y – 8x = 4 2y = 8x + 4 y = 4x + 2 x-intercept 0 = 4x + 2 -2 = 4x -1/2 = x y-intercept y = 4(0) +2 y = 2

    23. 2-7 Linear Inequalities Use “test points” to determine which side of the line should be shaded. (2y – 8x = 4) (-2, 2) 2(2) – 8(-2) = 4 4 – (-16) = 4 20 = 4 ? true (0, 0) 2(0) – 8(0) = 4 0 – 0 = 4 0 = 4 ? false So we shade the side of the line that includes the “true” point, (-2, 2)

    24. 2-7 Linear Inequalities Example: Graph 12 < -3y – 9x Graph the line. 12 ? -3y – 9x 3y ? -9x – 12 y ? -3x – 4 x-intercept 0 = -3x – 4 4 = -3x -4/3 = x y-intercept y = 3(0) – 4 y = -4

    25. 2-7 Linear Inequalities Use “test points” to determine which side of the line should be shaded. (12 < -3y – 9x) (-3, -3) 12 < -3(-3) – 9(-3) 12 < 9 + 27 12 < 36 ? true (0, 0) 12 < -3(0) – 9(0) 12 < 0 – 0 12 < 0 ? false So we shade the side of the line that includes the “true” point, (-3, -3)

    26. 2-7 Problems Graph each inequality. 2x > y – 4 e. 2y = 6|x| 5 = y f. 42x > 7y 4 < -2y g. |x| < y + 2 y = |x| + 3 h. x – 4 = 8y

    27. 2-7 Answers 1)a) b) c) d)

    28. 2-7 Answers 1)e) f) g) h)

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