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Graphing Equations and Inequalities

Learn how to graph linear equations, find intercepts, understand slope, equations of lines, and more in this detailed guide. Practice plotting points and solving equations in the rectangular coordinate system.

ortizmichelle
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Graphing Equations and Inequalities

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  1. Chapter 10 Graphing Equations and Inequalities

  2. Chapter Sections 10.1 – Reading Graphs and the Rectangular Coordinate System 10.2 – Graphing Linear Equations 10.3 – Intercepts 10.4 – Slope and Rate of Change 10.5 – Equations of Lines 10.6 – Introduction to Functions 10.7 – Graphing Linear Inequalities in Two Variables 10.8 – Direct and Inverse Variation

  3. Reading Graphs and the Rectangular Coordinate System § 10.1

  4. Vocabulary Ordered pair – a sequence of 2 numbers where the order of the numbers is important Axis – horizontal or vertical number line Origin – point of intersection of two axes Quadrants – regions created by intersection of 2 axes Location of a pointresiding in the rectangular coordinate system created by a horizontal (x-) axis and vertical (y-) axis can be described by an ordered pair. Each number in the ordered pair is referred to as a coordinate

  5. To graph the point corresponding to a particular ordered pair (a,b), you must start at the origin and move a units to the left of right (right if a is positive, left if a is negative), then move b units up or down (up if b is positive, down if b is negative). Graphing an Ordered Pair

  6. y-axis Quadrant I Quadrant II (0, 5) (5, 3) (-4, 2) 3 units up (0, 0) x-axis 5 units right (-6, 0) origin (2, -4) Quadrant IV Quadrant III Graphing an Ordered Pair Note that the order of the coordinates is very important, since (-4, 2) and (2, -4) are located in different positions.

  7. Vocabulary Paired data are data that can be represented as an ordered pair A scatter diagram is the graph of paired data as points in the rectangular coordinate system An order pair is a solution of an equation in two variables if replacing the variables by the appropriate values of the ordered pair results in a true statement.

  8. Solutions of an Equation Example Determine whether (3, –2) is a solution of 2x + 5y = –4. Let x = 3 and y = –2 in the equation. 2x + 5y = –4 2(3) + 5(–2) = –4 (replace x with 3 and y with –2) 6 + (–10) = –4 (compute the products) –4 = –4 (True) So (3, –2) is a solution of 2x + 5y = –4

  9. Solutions of an Equation Example Determine whether (–1, 6) is a solution of 3x–y = 5. Let x = –1 and y = 6 in the equation. 3x–y = 5 3(–1) – 6 = 5 (replace x with –1 and y with 6) –3 – 6 = 5 (compute the product) –9 = 5 (False) So (–1, 6) is not a solution of 3x – y = 5

  10. If you know one coordinate in an ordered pair that is a solution for an equation, you can find the other coordinate through substitution and solving the resulting equation. Solving an Equation

  11. –8 + 8 + 4y = 4 + 8 (add 8 to both sides) Solving an Equation Example Complete the ordered pair (4, ) so that it is a solution of –2x + 4y = 4. Let x = 4 in the equation and solve for y. –2x + 4y = 4 –2(4) + 4y = 4 (replace x with 4) –8 + 4y = 4 (compute the product) So the completed ordered pair is (4, 3). 4y = 12 (simplify both sides) y = 3 (divide both sides by 4)

  12. 4x + 2 – 2 = 4 – 2 (subtract 2 from both sides) Solving an Equation Example Complete the ordered pair (__, – 2) so that it is a solution of 4x – y = 4. Let y = – 2 in the equation and solve for x. 4x – y = 4 4x– (– 2) = 4 (replace y with – 2) 4x + 2 = 4 (simplify left side) So the completed ordered pair is (½, – 2). 4x = 2 (simplify both sides) x = ½ (divide both sides by 4)

  13. Graphing Linear Equations § 10.2

  14. Linear Equations Linear Equation in Two Variables Ax + By = C A, B, C are real numbers, A and B not both 0 This is called “standard form” Graphing Linear Equations Find at least 2 points on the line y = mx + b crosses the y-axis at b (called “slope-intercept form”)

  15. We find two ordered pair solutions (and a third solution as a check on our computations) by choosing a value for one of the variables, x or y, then solving for the other variable. We plot the solution points, then draw the line containing the 3 points. Graphing Linear Equations Example Graph the linear equation 2x–y = –4. Continued.

  16. Graphing Linear Equations Example continued Graph the linear equation 2x – y = – 4. Let x = 1. Then 2x – y = – 4 becomes 2(1) – y = – 4 (replace x with 1) 2 – y = – 4 (simplify the left side) – y = – 4 – 2 = – 6 (subtract 2 from both sides) y = 6 (multiply both sides by – 1) So one solution is (1, 6) Continued.

  17. Graphing Linear Equations Example continued Graph the linear equation 2x – y = – 4. For the second solution, let y = 4. Then 2x – y = – 4 becomes 2x – 4 = – 4 (replace y with 4) 2x = – 4 + 4 (add 4 to both sides) 2x = 0 (simplify the right side) x = 0 (divide both sides by 2) So the second solution is (0, 4) Continued.

  18. Graphing Linear Equations Example continued Graph the linear equation 2x – y = – 4. For the third solution, let x = – 3. Then 2x – y = – 4 becomes 2(– 3) – y = – 4 (replace x with – 3) – 6 – y = – 4 (simplify the left side) – y = – 4 + 6 = 2 (add 6 to both sides) y = – 2 (multiply both sides by – 1) So the third solution is (– 3, – 2) Continued.

  19. y (1, 6) (0, 4) x (– 3, – 2) Graphing Linear Equations Example continued Now we plot all three of the solutions (1, 6), (0, 4) and (– 3, – 2). And then we draw the line that contains the three points.

  20. Graph the linear equation y = x + 3. Since the equation is solved for y, we should choose values for x. To avoid fractions, we should select values of x that are multiples of 4 (the denominator of the fraction). Graphing Linear Equations Example Continued.

  21. Graph the linear equation y = x + 3. Then y = x + 3 becomes y = (4) + 3 (replace x with 4) Graphing Linear Equations Example continued Let x = 4. y = 3 + 3 = 6 (simplify the right side) So one solution is (4, 6) Continued.

  22. Graph the linear equation y = x + 3. Then y = x + 3 becomes y = (0) + 3 (replace x with 0) Graphing Linear Equations Example continued For the second solution, let x = 0. y = 0 + 3 = 3 (simplify the right side) So a second solution is (0, 3) Continued.

  23. Graph the linear equation y = x + 3. Then y = x + 3 becomes y = (– 4) + 3 (replace x with – 4) Graphing Linear Equations Example continued For the third solution, let x = – 4. y = – 3 + 3 = 0 (simplify the right side) So the third solution is (– 4, 0) Continued.

  24. y (4, 6) (0, 3) x (– 4, 0) Graphing Linear Equations Example continued Now we plot all three of the solutions (4, 6), (0, 3) and (– 4, 0). And then we draw the line that contains the three points.

  25. Intercepts § 10.3

  26. Intercepts Intercepts of axes (where graph crosses the axes) Since all points on the x-axis have a y-coordinate of 0, to find x-intercept, let y = 0 and solve for x Since all points on the y-axis have an x-coordinate of 0, to find y-intercept, let x = 0 and solve for y

  27. Intercepts = y(divide both sides by -3) So the y-intercept is (0, ) Example Find the y-intercept of 4 = x – 3y Let x = 0. Then 4 = x – 3y becomes 4 = 0 – 3y(replace x with 0) 4 = – 3y (simplify the right side)

  28. Intercepts Example Find the x-intercept of 4 = x – 3y Let y = 0. Then 4 = x – 3y becomes 4 = x – 3(0) (replace y with 0) 4 = x (simplify the right side) So the x-intercept is (4,0)

  29. Graph by Plotting Intercepts We previously found that the y-intercept is (0, ) and the x-intercept is (4, 0). Example Graph the linear equation 4 = x – 3y by plotting intercepts. Plot both of these points and then draw the line through the 2 points. Note: You should still find a 3rd solution to check your computations. Continued.

  30. Graph by Plotting Intercepts Example continued Graph the linear equation 4 = x – 3y. Along with the intercepts, for the third solution, let y = 1. Then 4 = x – 3y becomes 4 = x – 3(1)(replace y with 1) 4 = x – 3(simplify the right side) 4 + 3 = x(add 3 to both sides) 7 = x(simplify the left side) So the third solution is (7, 1) Continued.

  31. y Now we plot the two intercepts (0, ) and (4, 0) along with the third solution (7, 1). (0, ) (7, 1) (4, 0) x Graph by Plotting Intercepts Example continued And then we draw the line that contains the three points.

  32. Graph by Plotting Intercepts Example Graph 2x = y by plotting intercepts To find the y-intercept, let x = 0 2(0) = y 0 = y, so the y-intercept is (0,0) To find the x-intercept, let y = 0 2x = 0 x = 0, so the x-intercept is (0,0) Oops! It’s the same point. What do we do? Continued.

  33. Graph by Plotting Intercepts Example continued Since we need at least 2 points to graph a line, we will have to find at least one more point Let x = 3 2(3) = y 6 = y, so another point is (3, 6) Let y = 4 2x = 4 x = 2, so another point is (2, 4) Continued.

  34. y (3, 6) (2, 4) x (0, 0) Graph by Plotting Intercepts Example continued Now we plot all three of the solutions (0, 0), (3, 6) and (2, 4). And then we draw the line that contains the three points.

  35. Graph by Plotting Intercepts Example Graph y = 3 Note that this line can be written as y = 0·x + 3 The y-intercept is (0, 3), but there is no x-intercept! (Since an x-intercept would be found by letting y = 0, and 0  0·x + 3, there is no x-intercept) Every value we substitute for x gives a y-coordinate of 3 The graph will be a horizontal line through the point (0,3) on the y-axis Continued.

  36. y (0, 3) x Graph by Plotting Intercepts Example continued

  37. Graph by Plotting Intercepts Example Graph x = – 3 This equation can be written x = 0·y – 3 When y = 0, x = – 3, so the x-intercept is (– 3,0), but there is no y-intercept Any value we substitute for y gives an x-coordinate of –3 So the graph will be a vertical line through the point (– 3,0) on the x-axis Continued.

  38. y x (-3, 0) Graph by Plotting Intercepts Example continued

  39. Vertical and Horizontal Lines Vertical lines Appear in the form of x = c, where c is a real number x-intercept is at (c, 0), no y-intercept unless c = 0 (y-axis) Horizontal lines Appear in the form of y = c, where c is a real number y-intercept is at (0, c), no x-intercept unless c = 0 (x-axis)

  40. Slope and Rate of Change § 10.4

  41. Slope Slope of a Line

  42. Slope Example Find the slope of the line through (4, -3) and (2, 2) If we let (x1, y1) be (4, -3) and (x2, y2) be (2, 2), then Note:If we let (x1, y1) be (2, 2) and (x2, y2) be (4, -3), then we get the same result.

  43. Slope of a Horizontal Line For any 2 points, the y values will be equal to the same real number. The numerator in the slope formula = 0 (the difference of the y-coordinates), but the denominator  0 (two different points would have two different x-coordinates). So the slope = 0.

  44. Slope of a Vertical Line For any 2 points, the x values will be equal to the same real number. The denominator in the slope formula = 0 (the difference of the x-coordinates), but the numerator  0 (two different points would have two different y-coordinates), So the slope is undefined (since you can’t divide by 0).

  45. Summary of Slope of Lines If a line moves up as it moves from left to right, the slope is positive. If a line moves down as it moves from left to right, the slope is negative. Horizontal lines have a slope of 0. Vertical lines have undefined slope (or no slope).

  46. Parallel Lines Two lines that never intersect are called parallel lines. Parallel lines have the same slope • unless they are vertical lines, which have no slope. Vertical lines are also parallel.

  47. Parallel Lines Example Find the slope of a line parallel to the line passing through (0,3) and (6,0) So the slope of any parallel line is also –½

  48. Perpendicular Lines Two lines that intersect at right angles are called perpendicular lines Two nonvertical perpendicular lines have slopes that are negative reciprocals of each other The product of their slopes will be –1 Horizontal and vertical lines are perpendicular to each other

  49. Perpendicular Lines So the slope of any perpendicular line is Example Find the slope of a line perpendicular to the line passing through (1,3) and (2,-8)

  50. Parallel and Perpendicular Lines y = x + 1 (divide both sides by 5) The first equation has a slope of 5 and the second equation has a slope of , so the lines are perpendicular. Example Determine whether the following lines are parallel, perpendicular, or neither. 5x + y = 6 and x + 5y = 5 First, we need to solve both equations for y. In the first equation, y = 5x – 6 (add 5x to both sides) In the second equation, 5y = x + 5 (subtract x from both sides)

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