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

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 MTH 55

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 True  Since 18 = 18 is true, the pair (2, 3) is a solution

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  False

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

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.

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.

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

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)

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

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)

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 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

MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 13 Bruce Mayer, PE Chabot College Mathematics FedEx Soln: c = 2.8w  Select values for w and then calculate c.  c = 2.8w If w = 2, then c = 2.8(2) = If w = 4, then c = 2.8(4) = If w = 8, then c = 2.8(8) =  Tabulating the Results: wc

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

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

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)

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)

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.

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 y = 10 2y = 10 y = 5  Thus The y-intercept is (0, 5)

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 = 10 5x = 10 x = 2  Thus The x-intercept is (2, 0)

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 – y = 10 – y = 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)

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)

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)

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

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

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)

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

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)

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)

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

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

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

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

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

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)

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.

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

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)

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

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

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

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

MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt 43 Bruce Mayer, PE Chabot College Mathematics P  Find Slope for Lines  Recall

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

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

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