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Dec.18

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Dec.18

1. 1. Your double agent is lost in the woods. You havelocated her on the grid at (-2,-3). The enemy isclosing in fast, and the only two Safe houses arelocated in the direction of a -2/3 slope from herpresent location.Help your agent narrowly escape death by sendingher the coordinates of the two safe houses.
2. 2. 1. What is 250% of 18? 2. 24 is reduced by 25%; what number could you multiply 24by to get the answer? 3. Name three solutions to the equation y = -4x+1 4. Find the x and y intercepts, and the slope, then graph theequation 2x = 3y -6 5. Find the slope of the line containing the following points: (0,5) (3,7) (6,9) (9,11)6. The slope of the line between (2,10) and (x,4) is -3. Find the value of x.7. 7|x−10|−9 = 54
3. 3. Views of a Function
4. 4. Domain & Range
5. 5. What is y when x is 4?
6. 6. What is x when y is 4?
7. 7. How does y change as x increases?
8. 8. Graph the following equation: y = -3x - 7
9. 9. Mondays Topic: Graphing Equations with one variableAs you know, every point on a coordinate plane is theintersection of two variables, the independent x, andthe dependent y.What if, however, your equation only has anindependent x variable? 2x - 2 = -4These equations can still be solved by finding the xintercept, since we know that at this point, y = 0
10. 10. How is it Done? Task: Graph the equation 2x - 2 = - 4 1. First, set the equation equal to zero: 2x + 2 = 0 2. Replace 0 with f(x) 3. Make a table 4. Graph the ordered pairs Graph the Equation** Graph 5x + 2 = 7
11. 11. Todays Topic: Using the Slope- Intercept Form
12. 12. Finding the Equation of a Line
13. 13. 11-3 Using Slopes and InterceptsCourse 3
14. 14. 11-3 Using Slopes and Intercepts Learn to use slopes and intercepts to graph linear equations.Course 3
15. 15. 11-3 Insert Lesson Title Here As you watch the video, take notes on your handout.Course 3
16. 16. 11-3 Using Intercepts One way to graph a linear equation easily is by finding the x-intercept and the y-intercept. The x-intercept is the value of x where the line crosses the x-axis (y = 0). The y-intercept is the value of y where the line crosses the y-axis (x = 0).Course 3
17. 17. 11-3 Using Intercepts Example 1 Find the x-intercept and y-intercept of the line 4x – 3y = 12. Use the intercepts to graph the equation. Find the x-intercept (y = 0). 4x – 3y = 12 4x – 3(0) = 12 4x = 12 4x 12 = 4 4 x=3 The x-intercept is 3.Course 3
18. 18. 11-3 Using Intercepts Example 1 Continued Find the y-intercept (x = 0). 4x – 3y = 12 4(0) – 3y = 12 –3y = 12 -3y= 12 -3 -3 y = –4 The y-intercept is –4.Course 3
19. 19. 11-3 Using Intercepts 4x – 3y = 12 Crosses the x-axis at the point (3, 0) Crosses the y-axis at the point (0, –4)Course 3
20. 20. 11-3 Using Intercepts Try This Find the x-intercept and y-intercept of the line 8x – 6y = 48. Use the intercepts to graph the equation. Find the x-intercept (y = 0). 8x – 6y = 48 8x – 6(0) = 48 8x = 48 8x 48 = 8 8 x=6 The x-intercept is 6 so the point is (6, 0).Course 3
21. 21. 11-3 Using Intercepts Try This Find the y-intercept (x = 0). 8x – 6y = 48 8(0) – 6y = 48 –6y = 48 -6y= 48 -6 -6 y = –8 The y-intercept is –6 so the point is (0, -8).Course 3
22. 22. 11-3 Using Slopes and Intercepts Try This: Example 1 Continued The graph of 8x – 6y = 48 is the line that crosses the x-axis at the point (6, 0) and the y-axis at the point (0, –8).Course 3
23. 23. 11-3 Using Slopes and Intercepts In an equation written in slope-intercept form, y = mx + b, m is the slope and b is the y-intercept. y = mx + b Slope y-interceptCourse 3
24. 24. 11-3 Using Slopes and Intercepts Using the Slope-Intercept Form 1. Isolate the y so that you equation is in y = mx+b form. 2. Slope will always be “m” (the number in front of x). 3. The y-intercept will always be “b” (the number by itself). Write this point as (0,b) Helpful Hint For an equation such as y = x – 6, write it as y = x + (–6) to read the y-intercept, –6. The point would be (0,-6)Course 3
25. 25. 11-3 Using Slopes and Intercepts Example 2 Write each equation in slope-intercept form, and then find the slope and y- intercept. A. 2x + y = 3 2x + –2x –2x Subtract 2x from both sides. y = 3 – 2x y = –2x + 3 The equation is in slope-intercept form. The slope of the line is –2, m = –2 b = 3 and the y-intercept is 3.Course 3
26. 26. 11-3 Using Slopes and Intercepts More Examples B. 5y = 3x 5y = 3x 5y= 3 x Divide both sides by 5 to solve for y. 5 5 y =3 x + 0 The equation is in slope-intercept form. 5 m =3 b=0 5 The slope of the line is 3 , and the 5 y-intercept is 0.Course 3
27. 27. 11-3 Using Slopes and Intercepts More Examples C. 4x + 3y = 9 4x + 3y = 9 –4x –4x Subtract 4x from both sides. 3y = –4x + 9 3y= –4x +9 Divide both sides by 3. 3 3 3 4 The equation is in slope-intercept form. y =- 3x + 3 The slope of the line 4x+ 3y = 9 m =- 4 b = 3 is – 4 , and the y-intercept is 3. 3 3Course 3
28. 28. 11-3 Using Slopes and Intercepts Try This Write each equation in slope-intercept form, and then find the slope and y- intercept. A. –4x + y = 4 4x –4x Subtract 4x from both sides. y = 4 – 4x Rewrite to match slope-intercept form. y = –4x + 4 The equation is in slope-intercept form. m = –4 b = 4 The slope of the line 4x + y = 4 is –4, and the y-intercept is 4.Course 3
29. 29. 11-3 Using Slopes and Intercepts Try This B. 7y = 2x 7y = 2x 7y= 2 x Divide both sides by 7 to solve for y. 7 7 y =2 x + 0 The equation is in slope-intercept form. 7 m =2 b=0 7 2 The slope of the line 7y = 2x is 7 , and the y-intercept is 0.Course 3
30. 30. 11-3 Using Slopes and Intercepts Try This C. 5x + 4y = 8 5x + 4y = 8 –5x –5x Subtract 5x from both sides. 4y = 8 – 5x Rewrite to match slope-intercept form. 5x + 4y = 8 4y= –5x +8 Divide both sides by 4. 4 4 4 5 y =- 4x + 2 The equation is in slope-intercept form. The slope of the line 5x + 4y = 8 m =- 5 b = 2 is – 5 , and the y-intercept is 2. 4 4Course 3
31. 31. 11-3 Using Slopes and Intercepts Additional Example 3: Entertainment Application A video club charges \$8 to join, and \$1.25 for each DVD that is rented. The linear equation y = 1.25x + 8 represents the amount of money y spent after renting x DVDs. Graph the equation by first identifying the slope and y-intercept. The equation is in slope-intercept y = 1.25x + 8 form. m =1.25 b=8Course 3
32. 32. 11-3 Using Slopes and Intercepts Additional Example 3 Continued The slope of the line is 1.25, and the y-intercept is 8. The line crosses the y-axis at the point (0, 8) and moves up 1.25 units for every 1 unit it moves to the right.Course 3
33. 33. 11-3 Using Slopes and Intercepts Try This: Example 3 A salesperson receives a weekly salary of \$500 plus a commission of 5% for each sale. Total weekly pay is given by the equation S = 0.05c + 500. Graph the equation using the slope and y-intercept. The equation is in y = 0.05x + 500 slope-intercept form. m =0.05 b = 500Course 3
34. 34. 11-3 Using Slopes and Intercepts Try This: Example 3 Continued y The slope of the line is 0.05, and the y-intercept 2000 is 500. The line crosses 1500 the y-axis at the point (0, 500) and moves up 1000 0.05 units for every 1 unit 500 it moves to the right. x 5000 10,000 15,000Course 3
35. 35. 11-3 Using Slopes and Intercepts Additional Example 4: Writing Slope-Intercept Form Write the equation of the line that passes through (3, –4) and (–1, 4) in slope-intercept form. Find the slope. y2 – y1 x2 – x1 = 4 – (–4) = 8 = –2 The slope is –2. –1 – 3 –4 Choose either point and substitute it along with the slope into the slope-intercept form. y = mx + b Substitute –1 for x, 4 for y, and –2 4 = –2(–1) + b for m. 4=2+b Simplify.Course 3
36. 36. 11-3 Using Slopes and Intercepts Additional Example 4 Continued Solve for b. 4=2+b –2 –2 Subtract 2 from both sides. 2=b Write the equation of the line, using –2 for m and 2 for b. y = –2x + 2Course 3
37. 37. 11-3 Using Slopes and Intercepts Try This: Example 4 Write the equation of the line that passes through (1, 2) and (2, 6) in slope-intercept form. Find the slope. y2 – y1 x2 – x1 = 6–2 = 4 =4 The slope is 4. 2–1 1 Choose either point and substitute it along with the slope into the slope-intercept form. y = mx + b Substitute 1 for x, 2 for y, and 2 = 4(1) + b 4 for m. 2=4+b Simplify.Course 3
38. 38. 11-3 Using Slopes and Intercepts Try This: Example 4 Continued Solve for b. 2=4+b –4 –4 Subtract 4 from both sides. –2 = b Write the equation of the line, using 4 for m and –2 for b. y = 4x – 2Course 3
39. 39. 11-4 Point-Slope Form Warm Up Problem of the Day Lesson PresentationCourse 3
40. 40. 11-4 Point-Slope Form Warm Up Write the equation of the line that passes through each pair of points in slope-intercept form. 1. (0, –3) and (2, –3) y = –3 2. (5, –3) and (5, 1) x=5 3. (–6, 0) and (0, –2) y = – 1x – 2 3 4. (4, 6) and (–2, 0) y=x+2Course 3
41. 41. 11-4 Point-Slope Form Problem of the Day Without using equations for horizontal or vertical lines, write the equations of four lines that form a square. Possible answer: y = x + 2, y = x – 2, y = –x + 2, y = –x – 2Course 3
42. 42. 11-4 Point-Slope Form Learn to find the equation of a line given one point and the slope.Course 3
43. 43. Insert Lesson Title Here11-4 Point-Slope Form Vocabulary point-slope formCourse 3
44. 44. 11-4 Point-Slope Form The point-slope of an equation of a line with slope m passing through (x1, y1) is y – y1 = m(x – x1). Point on the line Point-slope form (x1, y1) y – y1 = m (x – x1) slopeCourse 3
45. 45. 11-4 Point-Slope Form Additional Example 1: Using Point-Slope Form to Identify Information About a Line Use the point-slope form of each equation to identify a point the line passes through and the slope of the line. A. y – 7 = 3(x – 4) y – y1 = m(x – x1) The equation is in point-slope y – 7 = 3(x – 4) form. Read the value of m from the m=3 equation. (x1, y1) = (4, 7) Read the point from the equation. The line defined by y – 7 = 3(x – 4) has slope 3, and passes through the point (4, 7).Course 3
46. 46. 11-4 Point-Slope Form Additional Example 1B: Using Point-Slope Form to Identify Information About a Line B. y – 1 = 1 (x + 6) 3 y – y1 = m(x – x1) y – 1 = 1 (x + 6) 3 1 Rewrite using subtraction y – 1 = 3 [x – (–6)] instead of addition. m =13 (x1, y1) = (–6, 1) The line defined by y – 1 = 1 (x + 6) has slope 1 , and 3 3 passes through the point (–6, 1).Course 3
47. 47. 11-4 Point-Slope Form Try This: Example 1 Use the point-slope form of each equation to identify a point the line passes through and the slope of the line. A. y – 5 = 2 (x – 2) y – y1 = m(x – x1) y – 5 = 2(x – 2) The equation is in point-slope form. Read the value of m from the m=2 equation. (x1, y1) = (2, 5) Read the point from the equation. The line defined by y – 5 = 2(x – 2) has slope 2, and passes through the point (2, 5).Course 3
48. 48. 11-4 Point-Slope Form Try This: Example 1B B. y – 2 = 2 (x + 3) 3 y – y1 = m(x – x1) y – 2 = 2 (x + 3) 3 y – 2 = 2 [x – (–3)] Rewrite using subtraction 3 instead of addition. m =2 3 (x1, y1) = (–3, 2) The line defined by y – 2 = 2 (x + 3) has slope 2 , and 3 3 passes through the point (–3, 2).Course 3
49. 49. 11-4 Point-Slope Form Additional Example 2: Writing the Point-Slope Form of an Equation Write the point-slope form of the equation with the given slope that passes through the indicated point. A. the line with slope 4 passing through (5, -2) y – y1 = m(x – x1) [y – (–2)] = 4(x – 5) Substitute 5 for x1, –2 for y1, and 4 for m. y + 2 = 4(x – 5) The equation of the line with slope 4 that passes through (5, –2) in point-slope form is y + 2 = 4(x – 5).Course 3
50. 50. 11-4 Point-Slope Form Additional Example 2: Writing the Point-Slope Form of an Equation B. the line with slope –5 passing through (–3, 7) y – y1 = m(x – x1) y – 7 = -5[x – (–3)] Substitute –3 for x1, 7 for y1, and –5 for m. y – 7 = –5(x + 3) The equation of the line with slope –5 that passes through (–3, 7) in point-slope form is y – 7 = –5(x + 3).Course 3
51. 51. 11-4 Point-Slope Form Try This: Example 2A Write the point-slope form of the equation with the given slope that passes through the indicated point. A. the line with slope 2 passing through (2, –2) y – y1 = m(x – x1) [y – (–2)] = 2(x – 2) Substitute 2 for x1, –2 for y1, and 2 for m. y + 2 = 2(x – 2) The equation of the line with slope 2 that passes through (2, –2) in point-slope form is y + 2 = 2(x – 2).Course 3
52. 52. 11-4 Point-Slope Form Try This: Example 2B B. the line with slope -4 passing through (-2, 5) y – y1 = m(x – x1) y – 5 = –4[x – (–2)] Substitute –2 for x1, 5 for y1, and –4 for m. y – 5 = –4(x + 2) The equation of the line with slope –4 that passes through (–2, 5) in point-slope form is y – 5 = –4(x + 2).Course 3
53. 53. 11-4 Point-Slope Form Additional Example 3: Entertainment Application A roller coaster starts by ascending 20 feet for every 30 feet it moves forward. The coaster starts at a point 18 feet above the ground. Write the equation of the line that the roller coaster travels along in point-slope form, and use it to determine the height of the coaster after traveling 150 feet forward. Assume that the roller coaster travels in a straight line for the first 150 feet. As x increases by 30, y increases by 20, so the slope 2 of the line is 20 or . The line passes through the 30 3 point (0, 18).Course 3
54. 54. 11-4 Point-Slope Form Additional Example 3 Continued y – y1 = m(x – x1) Substitute 0 for x1, 18 for y1, y – 18 = 2 (x – 0) 3 and 2 for m. 3 The equation of the line the roller coaster travels along, in point-slope form, is y – 18 = 2 x. Substitute 150 for x to find the value of y. 3 y – 18 = 2 (150) 3 y – 18 = 100 y = 118 The value of y is 118, so the roller coaster will be at a height of 118 feet after traveling 150 feet forward.Course 3
55. 55. 11-4 Point-Slope Form Try This: Example 3 A roller coaster starts by ascending 15 feet for every 45 feet it moves forward. The coaster starts at a point 15 feet above the ground. Write the equation of the line that the roller coaster travels along in point-slope form, and use it to determine the height of the coaster after traveling 300 feet forward. Assume that the roller coaster travels in a straight line for the first 300 feet. As x increases by 45, y increases by 15, so the slope 1 of the line is 15 or . The line passes through the 45 3 point (0, 15).Course 3
56. 56. 11-4 Point-Slope Form Try This: Example 3 Continued y – y1 = m(x – x1) Substitute 0 for x1, 15 for y1, y – 15 = 1 (x – 0) 3 and 1 for m. 3 The equation of the line the roller coaster travels 1 along, in point-slope form, is y – 15 = x. Substitute 300 for x to find the value of y. 3 y – 15 = 1 (300) 3 y – 15 = 100 y = 115 The value of y is 115, so the roller coaster will be at a height of 115 feet after traveling 300 feet forward.Course 3