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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics §2.4a Lines by Intercepts

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review § Any QUESTIONS About § 2.3 → Algebra of Funtions Any QUESTIONS About HomeWork § 2.2 → HW-05 2.3 MTH 55

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 3 Bruce Mayer, PE Chabot College Mathematics Eqn of a Line Ax + By = C Determine whether each of the following pairs is a solution of eqn 4y + 3x = 18: a) (2, 3); b) (1, 5). Soln-a) We substitute 2 for x and 3 for y 4y + 3x = 18 43 + 32 | 18 12 + 6 | 18 18 = 18 True Since 18 = 18 is true, the pair (2, 3) is a solution

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 4 Bruce Mayer, PE Chabot College Mathematics Example Eqn of a Line Soln-b) We substitute 1 for x and 5 for y Since 23 = 18 is false, the pair (1, 5) is not a solution 4y + 3x = 18 45 + 31 | 18 20 + 3 | 18 23 = 18 False

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 5 Bruce Mayer, PE Chabot College Mathematics To Graph a Linear Equation 1.Select a value for one coordinate and calculate the corresponding value of the other coordinate. Form an ordered pair. This pair is one solution of the equation. 2.Repeat step (1) to find a second ordered pair. A third ordered pair can be used as a check. 3.Plot the ordered pairs and draw a straight line passing through the points. The line represents ALL solutions of the equation

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 6 Bruce Mayer, PE Chabot College Mathematics Example Graph y = −4x + 1 Solution: Select convenient values for x and compute y, and form an ordered pair. If x = 2, then y = −4(2)+ 1 = −7 so (2,−7) is a solution If x = 0, then y = −4(0) + 1 = 1 so (0, 1) is a solution If x = –2, then y = −4(−2) + 1 = 9 so (−2, 9) is a solution.

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 7 Bruce Mayer, PE Chabot College Mathematics Example Graph y = −4x + 1 Results are often listed in a table. xy(x, y) 2–7–7(2, –7) 01(0, 1) –2–29(–2, 9) Choose x Compute y. Form the pair (x, y). Plot the points.

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 8 Bruce Mayer, PE Chabot College Mathematics Example Graph y = −4x + 1 Note that all three points line up. If they didn’t we would know that we had made a mistake Finally, use a ruler or other straightedge to draw a line Every point on the line represents a solution of: y = −4x + 1

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 9 Bruce Mayer, PE Chabot College Mathematics Example Graph x + 2y = 6 Solution: Select some convenient x-values and compute y-values. If x = 6, then 6 + 2y = 6, so y = 0 If x = 0, then 0 + 2y = 6, so y = 3 If x = 2, then 2 + 2y = 6, so y = 2 In Table Form, Then Plotting xy(x, y) 60(6, 0) 03(0, 3) 22(2, 2)

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 10 Bruce Mayer, PE Chabot College Mathematics Example Graph 4y = 3x Solution: Begin by solving for y. Or y is 75% of x To graph the last Equation we can select values of x that are multiples of 4 This will allow us to avoid fractions when computing the corresponding y-values

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 11 Bruce Mayer, PE Chabot College Mathematics Example Graph 4y = 3x Solution: Select some convenient x-values and compute y-values. If x = 0, then y = ¾ (0) = 0 If x = 4, then y = ¾ (4) = 3 If x = −4, then y = ¾ (−4) = −3 In Table Form, Then Plotting xy(x, y) 00(0, 0) 43(4, 3) −4−4−3−3 ( 4, 3)

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 12 Bruce Mayer, PE Chabot College Mathematics Example Application The cost c, in dollars, of shipping a FedEx Priority Overnight package weighing 1 lb or more a distance of 1001 to 1400 mi is given by c = 2.8w + 21.05 where w is the package’s weight in lbs Graph the equation and then use the graph to estimate the cost of shipping a 10½ pound package

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 13 Bruce Mayer, PE Chabot College Mathematics FedEx Soln: c = 2.8w + 21.05 Select values for w and then calculate c. c = 2.8w + 21.05 If w = 2, then c = 2.8(2) + 21.05 = 26.65 If w = 4, then c = 2.8(4) + 21.05 = 32.25 If w = 8, then c = 2.8(8) + 21.05 = 43.45 Tabulating the Results: wc 226.65 432.25 843.45

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 14 Bruce Mayer, PE Chabot College Mathematics FedEx Soln: Graph Eqn Plot the points. Weight (in pounds) Mail cost (in dollars) To estimate costs for a 10½ pound package, we locate the point on the line that is above 10½ lbs and then find the value on the c-axis that corresponds to that point 10 ½ pounds The cost of shipping an 10½ pound package is about $51.00 $51

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 15 Bruce Mayer, PE Chabot College Mathematics Finding Intercepts of Lines An “Intercept” is the point at which a line or curve, crosses either the X or Y Axes A line with eqn Ax + By = C (A & B ≠ 0) will cross BOTH the x-axis and y-axis The x-CoOrd of the point where the line intersects the x-axis is called the x-intercept The y-CoOrd of the point where the line intersects the y-axis is called the y-intercept

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 16 Bruce Mayer, PE Chabot College Mathematics Example Axes Intercepts For the graph shown a) find the coordinates of any x-intercepts b) find the coordinates of any y-intercepts Solution a) The x-intercepts are (−2, 0) and (2, 0) b) The y-intercept is (0,−4)

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 17 Bruce Mayer, PE Chabot College Mathematics Graph Ax + By = C Using Intercepts 1.Find the x-Intercept Let y = 0, then solve for x 2.Find the y-Intercept Let x = 0, then solve for y 3.Construct a CheckPoint using any convenient value for x or y 4.Graph the Equation by drawing a line thru the 3-points (i.e., connect the dots)

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 18 Bruce Mayer, PE Chabot College Mathematics To FIND the Intercepts To find the y-intercept(s) of an equation’s graph, replace x with 0 and solve for y. To find the x-intercept(s) of an equation’s graph, replace y with 0 and solve for x.

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 19 Bruce Mayer, PE Chabot College Mathematics Example Find Intercepts Find the y-intercept and the x-intercept of the graph of 5x + 2y = 10 SOLUTION: To find the y-intercept, we let x = 0 and solve for y 5 0 + 2y = 10 2y = 10 y = 5 Thus The y-intercept is (0, 5)

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 20 Bruce Mayer, PE Chabot College Mathematics Example Find Intercepts cont. Find the y-intercept and the x-intercept of the graph of 5x + 2y = 10 SOLUTION: To find the x-intercept, we let y = 0 and solve for x 5x + 2 0 = 10 5x = 10 x = 2 Thus The x-intercept is (2, 0)

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 21 Bruce Mayer, PE Chabot College Mathematics Example Graph w/ Intercepts Graph 5x + 2y = 10 using intercepts SOLUTION: We found the intercepts in the previous example. Before drawing the line, we plot a third point as a check. If we let x = 4, then –5 4 + 2y = 10 – 20 + 2y = 10 – 2y = −10 – y = − 5 We plot Intercepts (0, 5) & (2, 0), and also (4,−5) 5x + 2y = 10 x-intercept (2, 0) y-intercept (0, 5) Chk-Pt (4,-5)

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 22 Bruce Mayer, PE Chabot College Mathematics Example Graph w/ Intercepts Graph 3x − 4y = 8 using intercepts SOLUTION: To find the y-intercept, we let x = 0. This amounts to ignoring the x-term and then solving. −4y = 8 y = −2 Thus The y-intercept is (0, −2)

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 23 Bruce Mayer, PE Chabot College Mathematics Example Graph w/ Intercepts Graph 3x – 4y = 8 using intercepts SOLUTION: To find the x-intercept, we let y = 0. This amounts to ignoring the y-term and then solving 3x = 8 x = 8/3 Thus The x-intercept is (8/3, 0)

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 24 Bruce Mayer, PE Chabot College Mathematics Example Graph w/ Intercepts Construct Graph for 3x – 4y = 8 Find a third point. If we let x = 4, then – 34 – 4y = 8 – 12 – 4y = 8 – –4y = –4 – y = 1 We plot (0, −2), (8/3, 0), and (4, 1) and Connect the Dots x-intercept y-intercept 3x 4y = 8 Chk-Pt Charlie

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 25 Bruce Mayer, PE Chabot College Mathematics Example Graph y = 2 SOLUTION: We regard the equation y = 2 as the equivalent eqn: 0x + y = 2. No matter what number we choose for x, we find that y must equal 2. (−4, 2)2−4−4 (4, 2)24 (0, 2)20 (x, y)yx Choose any number for x. y must be 2. y=2

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 26 Bruce Mayer, PE Chabot College Mathematics Example Graph y = 2 Next plot the ordered pairs (0, 2), (4, 2) & (−4, 2) and connect the points to obtain a horizontal line. Any ordered pair of the form (x, 2) is a solution, so the line is parallel to the x-axis with y-intercept (0, 2) y = 2 ( 4, 2) (0, 2) (4, 2)

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 27 Bruce Mayer, PE Chabot College Mathematics Example Graph x = −2 SOLUTION: We regard the equation x = −2 as x + 0y = −2. We build a table with all −2’s in the x-column. xy(x, y) −2−24(−2, 4) −2−21(−2, 1) −2−2−4−4(−2, −4) x must be 2. Any number can be used for y. x = −2

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 28 Bruce Mayer, PE Chabot College Mathematics Example Graph x = −2 When we plot the ordered pairs (−2,4), (−2,1) & (−2, −4) and connect them, we obtain a vertical line Any ordered pair of the form (−2,y) is a solution. The line is parallel to the y-axis with x-intercept (−2,0) x = 2 ( 2, 4) ( 2, 4) ( 2, 1)

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 29 Bruce Mayer, PE Chabot College Mathematics Linear Eqns of ONE Variable The Graph of y = b is a Horizontal Line, with y-intercept (0,b) The Graph of x = a is a Vertical Line, with x-intercept (a,0)

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 30 Bruce Mayer, PE Chabot College Mathematics Example Horiz or Vert Line Write an equation for the graph SOLUTION: Note that every point on the horizontal line passing through (0,−3) has −3 as the y-coordinate. Thus The equation of the line is y = −3

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 31 Bruce Mayer, PE Chabot College Mathematics Example Horiz or Vert Line Write an equation for the graph SOLUTION: Note that every point on the vertical line passing through (4, 0) has 4 as the x-coordinate. Thus The equation of the line is x = 4

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 32 Bruce Mayer, PE Chabot College Mathematics SLOPE Defined SLOPE The SLOPE, m, of the line containing points (x 1, y 1 ) and (x 2, y 2 ) is given by

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 33 Bruce Mayer, PE Chabot College Mathematics Example Slope City Graph the line containing the points (−4, 5) and (4, −1) & find the slope, m SOLUTION Thus Slope m = −3/4 Change in y = − 6 Change in x = 8

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 34 Bruce Mayer, PE Chabot College Mathematics Example ZERO Slope Find the slope of the line y = 3 ( 3, 3) (2, 3) SOLUTION: Find Two Pts on the Line Then the Slope, m A Horizontal Line has ZERO Slope

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 35 Bruce Mayer, PE Chabot College Mathematics Example UNdefined Slope Find the slope of the line x = 2 SOLUTION: Find Two Pts on the Line Then the Slope, m A Vertical Line has an UNDEFINED Slope (2, 4) (2, 2)

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 36 Bruce Mayer, PE Chabot College Mathematics Applications of Slope = Grade Some applications use slope to measure the steepness. For example, numbers like 2%, 3%, and 6% are often used to represent the grade of a road, a measure of a road’s steepness. That is, a 3% grade means that for every horizontal distance of 100 ft, the road rises or falls 3 ft.

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 37 Bruce Mayer, PE Chabot College Mathematics Grade Example Find the slope (or grade) of the treadmill 0.42 ft 5.5 ft SOLUTION: Noting the Rise & Run In %-Grade for Treadmill

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 38 Bruce Mayer, PE Chabot College Mathematics Slope Symmetry We can Call EITHER Point No.1 or No.2 and Get the Same Slope Example, LET (x 1,y 1 ) = (−4,5) Moving L→R (−4,5) Pt1 (4,−1)

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 39 Bruce Mayer, PE Chabot College Mathematics Slope Symmetry cont Now LET (x 1,y 1 ) = (4,−1) (−4,5) (4,−1) Pt1 Moving R→L Thus

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 40 Bruce Mayer, PE Chabot College Mathematics Slopes Summarized POSITIVE Slope NEGATIVE Slope

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 41 Bruce Mayer, PE Chabot College Mathematics Slopes Summarized ZERO Slope UNDEFINED Slope slope = 0 Note that when a line is horizontal the slope is 0 slope = undefined Note that when the line is vertical the slope is undefined

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 42 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work Problems From §2.4 Exercise Set 26 (PPT), 12, 24, 52, 56 More Lines

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 43 Bruce Mayer, PE Chabot College Mathematics P2.4-26 Find Slope for Lines Recall

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 44 Bruce Mayer, PE Chabot College Mathematics All Done for Today Some Slope Calcs

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 45 Bruce Mayer, PE Chabot College Mathematics 20x20 Grid

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BMayer@ChabotCollege.edu MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 46 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics Appendix –

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