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Standard Form Ax + By = C Slopey-interceptx-intercept 3x – 4y = 24 –1/23 2x – 5y = –10 none – 7 3/4 – 6 8 x + 2y = 6 6 2/52 – 5 x = – 7 undefined.

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Presentation on theme: "Standard Form Ax + By = C Slopey-interceptx-intercept 3x – 4y = 24 –1/23 2x – 5y = –10 none – 7 3/4 – 6 8 x + 2y = 6 6 2/52 – 5 x = – 7 undefined."— Presentation transcript:

1 Standard Form Ax + By = C Slopey-interceptx-intercept 3x – 4y = 24 –1/23 2x – 5y = –10 none – 7 3/4 – 6 8 x + 2y = 6 6 2/52 – 5 x = – 7 undefined

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3 y – y 1 = m (x – x 1 ) m = slope x 1 = x-coordinate of point on line y 1 = y-coordinate of point on line To convert to Slope-intercept Form: DDistribute the slope. MMove y 1 to right side of the equal sign.

4 Example2/19/2016 Give an equation in Standard Form for the line that has a slope of passes through the point (8, –2). Find the x-int and y-int. Slope = – 2 and (8, –2) Point-Slope Form → Substitute your given values → Distribute → Continue →

5 Continued2/19/2016 Multiply by – 2 → Standard Form → Continued from the previous slide → Put into Standard Form → Solve for y → x-int = 28/3 y-int = –14

6 Example2/19/2016 Give an equation in Standard Form for the line that passes through the points (3, 5) and (6, – 1) Slope = – 2 Pick one of the given points. Let’s pick (3, 5) Point-Slope Form → Substitute your given values → Distribute → Solve for y → Put into Standard Form →

7 Example Continued 2/19/2016 Give an equation in Standard Form for the line that passes through the points (3, 5) and (6, – 1) Slope = – 2 Pick one of the given points. Let’s pick (6, –1) Point-Slope Form → Substitute your given values → Distribute → Solve for y → Put into Standard Form →

8 Example2/19/2016 Give an equation in Standard Form for the line that has an x-int 3 and y-int 7. Slope = – 7/3 Pick one of the given points. Let’s pick (3, 0) Point-Slope Form → Substitute your given values → Distribute → Put into Standard Form → The ordered pairs are (3, 0) and (0, 7). Continued →

9 Continued2/19/2016 Multiply by 3 → Standard Form → Continued from the previous slide →

10 Example2/19/2016 Last week, Mr. Baxter sold $20,000 worth of newspaper advertisements and earned $800. The week before, he sold $26,000 worth of advertisements and earned $860. Assume the relationship between Mr. Baxter’s weekly earnings and the value of the advertisements he sells is linear. a)Write the data above as two ordered pairs in the form (s, e). b) Find the slope of the line given these ordered pairs. c) Write an equation for the line in slope-intercept form given the data. (20,000, 800) and (26,000, 860) Slope =.01 Continued→

11 Example2/19/2016 Last week, Mr. Baxter sold $20,000 worth of newspaper advertisements and earned $800. The week before, he sold $26,000 worth of advertisements and earned $860. Assume the relationship between Mr. Baxter’s weekly earnings and the value of the advertisements he sells is linear. Ordered Pairs (20,000, 800) and (26,000, 860) Slope =.01 c) Write an equation for the line in slope-intercept form given the data. Slope =.01 and (20,000, 800)

12  You cannot find 1 equation for a piecewise function.  You can find an equation for each part.  Use the endpoints of each part to find the slope of that segment.  Use the slope and one of the points to write an equation in point-slope form.  Convert to slope intercept form.  Write an inequality to define that piece.

13  Use the endpoints of each part to find slope.  Write an equation in point-slope form.  Convert to slope- intercept form.  Write an inequality.

14  Use the endpoints of each part to find slope.  Write an equation in point-slope form.  Convert to slope- intercept form.  Write an inequality.

15  Use the endpoints of each part to find slope  write an equation in point-slope form  Convert to slope intercept  Write an inequality

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17  Lesson Master 3-5.  Pages 165-166 #’s 2, 5, 7-9, 13, 14, 16.  Quiz on Sections 3-4 and 3-5 on Tuesday, November 25 th !!!


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