1 / 43

Graph of Linear Equations

Graph of Linear Equations. Objective: Graph linear equations. The graph of any linear equation is a straight line.

blade
Télécharger la présentation

Graph of Linear Equations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Graph of Linear Equations • Objective: • Graph linear equations

  2. The graph of any linear equation is a straight line.

  3. Since two points determine a line, we can graph a linear equation by finding two points that belong to the graph. Then we draw a line containing those point. A third point should always be used as a check. Often the easiest points to find are the points where the graph crosses the axes. • The y-intercept of a graph is the y-coordinate of the point where the graph intersects the y-axis. (0,4) (5,0) The x-intercept is the x-coordinate of the points where the graph crosses the x-axis.

  4. Graph 4x + 5y = 20 • First find the intercepts. • To find the y-intercept, let x = 0 and solve for y. we find ____? y=4 We plot the point (0,4). To find the x-intercept, let y=0 and solve for x. we find _____? x=5 We plot the point (5,0).

  5. Theorem #2: The graph of y=mx is a line containing the origin. The graph of y=mx+b is a line parallel to y=mx and has b as the y-intercept. Y=2x Y=2x - b (0,b)

  6. Assignment:Graph and compare the graph of y=2x with 1. Y=2x+1 and y=2x-4

  7. Section #2:Objective: Graph the equation whose graphs are parallel to the x-axis or y-axis • Theorem #3. For constant a and b of an equation of the form y = b, is a line parallel to the x-axis with y intercept b. The graph of an equation of the form x = a is a line parallel to the y-axis with x-intercept a.

  8. Example: Any ordered pair (x,4) is a solution thus the line is parallel to the x-axis and the y-intercept is 4. Y (0,4) Y=4 X

  9. Exercise: Graph these equations. • X = 4 • Y = -3 • Y = 0

  10. Graphing Linear Equations: • 1. If there is a variable missing, solve for the other variable. The graph will be a line parallel to an axis. • 2. If no variable is missing, find the intercepts. Use the intercepts to graph. • 3. If the intercept points are too close together, or are the same point, choose another point farther the origin. • 4. Use a third point as a check.

  11. Section III. • Objective: • To find the slope of a line given two points on it. • To find the slopes of horizontal and vertical lines. • To find the point-slope form of the equation of a line.

  12. How will / can we find the slope of a line containing a given pair of points???

  13. When we say slope!? What comes to your mind?? • The ratio of the change in y to the change in x - slope of a line. DEFINITION: The slope m of a line is the change in y divided by the change in x or m = y2 – y1 / / x2-x1 Where (x1, y1) & (x2,y2) are any two points on the line & x2 is not equal to x1 We usually use the letter m to designate slope.

  14. Example: • The points (1,2) & (3,6) are on a line, find its slope

  15. What if we use the opposite order given example? Do we get the same slope?? • Illustrate show your solution.

  16. When we compute the slope, the order of the points DOES NOT matter as long as we take the same order of finding the difference

  17. The points (0,0) (-1,-2) are also on the line if we compute for the slope , we get the same answer, correct?? REMEMBER: if a line slants up from the left to right, it has a positive slope if a line slants down from left to right it has a negative slope. (3,6) (1,2) (-1,-2)

  18. m is positive number greater than 1 m=5 m is positive between 1 and 0 m=1 m=1/4

  19. Section #4:Objective: Find the slope of a horizontal & vertical line. Find the slope of the line y = 3?

  20. Vertical & Horizontal lines do not slant. Any two points on a horizontal line have the same y-coordinate. The change in y is 0, so the slope is 0. Try on your notebook. Find the slope of the line x = -4. Theorem #4 A horizontal line has slope 0. A vertical line has no slope.

  21. Exercise: Find the slope, if it exists. • Y = -5 • X = 17

  22. Section #5 • Objective: • Use the point-slope equation to find an equation of a line. If we know the slope of a line and the coordinates of a point on the line, we can find an equation of the line point – slope equation.

  23. Theorem #5.The Point Slope Equation • A line containing (x1,y1) with slope m has an equation (y- y1)=m (x- x1) Example: Find an equation of a line containing (1/2 , -1) with slope 5. (y- y1)=m (x- x1) (y –(-1)= 5(x – ½) y+1= 5(x – ½) y=5x-7/2

  24. Example: Find an equation of the line with y-intercept 4, with slope 3/5. (y- y1)=m (x- x1) y-4=3/5(x-0) y=3/5x + 4 Exercise: 1.Find an equation of the line containing the point (-2,4) with slope –3 2. Find an equation of the line containing the point (-4,-10) with slope ¼. 3. Find an equation of the line with x- intercept 5 and slope –1/2.

  25. Section #6 Objective: Given two points, we can find an equation of the line containing them. If we find the slope of a line dividing the change in y by change in x, and substitute this value for m in the point=slope equation, we obtain the two-point equation.

  26. Theorem #6 • The Two-Point Equation • Any non-vertical line tine containing the points (x1,y1) and (x2,y2) has an equation y - y1 = y2-y1 (x-x1) x2-x1

  27. Example: Find an equation of the line containing the points (2,3)and (1,-4), We find the slope and then substitute in the two-point equation. We take (2,3) as (x1,y1) and (1,-4) as (x2,y2). Sol’n: y-3 = -4 –3 (x-2) 1 –2 y = 7x –11

  28. Question: if we use the other pair, do we still get the same equation?? You try on your notebook!!! YES!!!

  29. Exercise: • Find an equation of the line containing the following pairs of points. • 1. (1,4) and (3,-2) • 2. (3,-6) and (0,4) • 3. (2,-5) and (7,1)

  30. Section #7 • Objective: Find the slope and y-intercept of a line, given the slope intercept equation for the line. Theorem #7 The Slope-Intercept Equation A non-vertical line with slope m and y-intercept b has an equation - y= mx+ b.

  31. Example: Find the slope and y-intercept of the line whose equation is y = 2x-3 y = 2x-3 Slope 2 y-intercept –3

  32. Exercise: 1.Find an equation of the line containing the point (21,9) with slope –1/32. Find an equation of the line containing the point (4,20) with slope ¼.3. Find an equation of the line with x- intercept 3 and slope 2. 4. Find an equation of the line containing the points (3,3)and (9,-4),5.Find an equation of the line containing the following pairs of points. 1. (5,4) and (3,-8) 2. (8,-6) and (0,5) 3. (7,-5) and (7,3) 6. Find the slope and y-intercept of the line whose equation is 9 = 1/4x+y 7. Find the slope and y-intercept of the line whose equation is -6y=-3x+7

  33. Section #8 • Graphing Using Slope-Intercept Form • Objective: Graph linear equation in slope-intercept form Example: Graph 5y – 20 = -3x So HOW are we going to solve for this??

  34. Solution • Solve for y, we find the slope-intercept form y = -3/5 x + 4. Thus, the y-intercept is 4 and the slope is –3/5

  35. DISTANCE FORMULA • The distance formula can be used to find the distance between two points when we know the coordinates of the points ANY IDEA WHAT IS THE DISTANCE FORMULA? REFRESH YOUR SECOND YEAR TOPIC?

  36. Theorem #10 • The Distance Formula The distance between any two points (x1,y1) and (x2,y2) is given by D = sqrt [(x1-x2)2 + (y1-y2) 2]

  37. Example: • Find the distance between points (8,7) and (3,-5) D = sqrt [(x1-x2)2 + (y1-y2) 2] D = sqrt [(8-3)2 + (7-(-5)) 2] D = sqrt (25 + 144 ) D = sqrt (169) D = 13

  38. Exercise: Find the distance between the points • (-5,3) and (2, -7) • (3, 3) and (-3, -3)

  39. MIDPOINT OF SEGMENTS Objective: Find the coordinates of the midpoint of a segment, given the coordinates of the endpoints. The coordinates of the midpoint of a segments can be found by averaging the coordinates of the endpoints. We can use the distance formula to verify a formula for finding the coordinates of the midpoint of a segment when the coordinates of the endpoints are known.

  40. Theorem #11 • The Midpoint Formula • If the coordinates of the endpoints of a segment are (x1,y1) and (x2,y2), then the coordinates of the midpoint are [(x1+ x2) / 2 , (y1+ y2) / 2 ]

  41. Example: • Find the coordinates of the midpoint of the segment with endpoints (-3, 5) and (4, -7) • Using the midpoint formula, we get • [(-3+ 4) / 2 , (5+(-7)) / 2 ] • (1/2 , -1)

  42. SEATWORK!!! IN 1 WHOLE PAD PAPER ANSWER THE FOLLOWING!

  43. 1.Find an equation of the line containing the point (21,9) with slope –1/3 2. Find an equation of the line containing the point (4,20) with slope ¼. 3. Find an equation of the line with x- intercept 3 and slope 2. 4. Find an equation of the line containing the points (3,3)and (9,4). 5.Find an equation of the line containing the following pairs of points.a. (5,4) and (3,-8) • b. (8,-6) and (0,5) • c. (7,-5) and (7,3) 6. Find the slope and y-intercept of the line whose equation is 9 = 1/4x+y 7. Find the slope and y-intercept of the line whose equation is -6y=-3x+7 8. Find the distance between the points a. (-7,3) and (2, -7) b. (2, 2) and (-2, -2) 9. Find the coordinates of the midpoint of the segment with endpoints (3, -5) and (4, -7)

More Related