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UNIT 1B Review of Linear Functions and Slope 1 Slope of a Line The term slope is often used to describe steepness or rate of change. The pitch of a roof, the steepness of a ski run, the speed of a car are all examples of slope. In each case, the slope is the ratio of the rise to the run In a coordinate system, we can determine the slope of any line segment from its endpoints P1(x1, y1) and P2(x2, y2). From P1 to P2: • the rise is the difference in the y-coordinates: y2 – y1 or ∆y • the run is the difference in the x-coordinates: x2 – x1 or ∆x 2 Slope of a Line Let 𝑃1 𝑥1 , 𝑦1 and 𝑃2 𝑥2 , 𝑦2 be points on a non vertical line, L. The slope of L is 𝒓𝒊𝒔𝒆 𝒚𝟐 − 𝒚𝟏 𝒎= = 𝒓𝒖𝒏 𝒙𝟐 − 𝒙𝟏 3 Slope of a Line A line that goes uphill as x increases has a positive slope.. A line that goes downhill as x increases has a negative slope. 4 Slope of a Line A horizontal line has slope zero since all of its points have the same y-coordinate. P1(-1, 1) P2(2, 1) 𝒎= 𝟏−𝟏 𝟎 = =𝟎 𝟐 − (−𝟏) 𝟑 For vertical lines, the slope is undefined since all of its points have the same x-coordinate 𝟐 − (−𝟏) 𝟑 𝒎= = 𝟏. 𝟓 − 𝟏. 𝟓 𝟎 P1 (1.5, 2) division by 0 is undefined P2 (1.5, -1) 5 Parallel Lines Parallel lines form equal angles with the x-axis. Hence, non-vertical parallel lines have the same slope. m1 = m2 6 Perpendicular Lines If two non-vertical lines L1 and L2 are perpendicular, their slopes m1 and m2 satisfy m1m2 = – 1, so each slope is the negative reciprocal of the other: 𝟑 Line 1 Slope = 𝟏 = 𝟑 3 1 × − = −1 1 3 𝟏 Line 2 Slope = − 𝟑 7 A(2, 1), B(5, 3) 𝟏 − 𝟑 −𝟐 𝟐 𝒎= = = 𝟐 − 𝟓 −𝟑 𝟑 Run = 3 Run = 3 Rise = 2 C(– 2, 2), D(1, 4) Rise = 2 𝟐−𝟒 −𝟐 𝟐 𝒎= = = −𝟐 − 𝟏 −𝟑 𝟑 Lines that are higher on the right have a positive slope. Lines that have equal slopes are parallel AB // CD and EF // GH 8 E(– 3 , 4), F(– 1 , – 2) 𝟒 − (−𝟐) 𝟔 𝒎= = = −𝟑 −𝟑 − (−𝟏) −𝟐 Run = – 1 Rise = 3 G(0, 5), H(1, 2) Rise = – 6 𝟓−𝟐 𝟑 𝒎= = = −𝟑 𝟎 − 𝟏 −𝟏 Run = 2 Lines that are higher on the left have a negative slope. Lines that have equal slopes are parallel AB // CD and EF // GH 9 J(3 , –2), K(6 , – 4) −𝟐 − (−𝟒) 𝟐 𝟐 𝒎= = =− 𝟑−𝟔 −𝟑 𝟑 L(4, –3), M(2, –6) Run = –3 −𝟑 − (−𝟔) 𝟑 𝒎= = 𝟒−𝟐 𝟐 Run = 2 Rise = 2 Rise = 3 𝟐 𝟑 − × = −𝟏 𝟑 𝟐 If the slopes of 2 lines are negative reciprocals (product = – 1) they are perpendicular. JK ┴ LM 10 R(– 2 , 4), S(5 , 4) 𝟒−𝟒 𝟎 𝒎= = =𝟎 −𝟐 − 𝟓 −𝟕 P(– 3, – 2 ), Q(– 3, 3) −𝟐 − 𝟑 −𝟔 𝒎= = −𝟑 − (−𝟑) 𝟎 ÷ by 0 is undefined Horizontal lines will always have a slope of zero Vertical lines will always have a slope which is undefined. They are also perpendicular 11 QUESTION 9: Two students entered a car rally. During part of the rally, they had to drive at a constant speed. The following graph shows the distance traveled over a given time while traveling at this constant speed. P2 What is the slope of the line and what does it represent? Distance (km) P1 Time (hours) Slope = (180 km – 60 km) = 60 km/h (3 h – 1 h) It represents the velocity of the car. 12 QUESTION 10 The pool at a fitness club is being drained. The graph shows the number of kilolitres of water remaining after an elapsed time. What isthe the intercept What slope of along thealong c)a)b) What isisthe intercept the theand vertical axis and whatthis line what does it does horizontal axis and what does this intercept represent? represent? intercept represent? The horizontal intercept is (400, 0) The vertical intercept is (0, 100) It takes 400 min to drain the pool. The pool has 100 kL in it before it is drained. Slope = (40 kL – 100kL) (240 min – 0 min) = – 0.25kL/min The pool is draining at a rate of 0.25 kL/min 13 Equations of Lines x y –4 3 –2 3 0 3 2 3 4 3 The horizontal line through the point (2, 3) has equation y = 3 x y 2 –4 2 –2 2 0 2 2 2 4 The vertical line through the point (2, 3) has equation x=2 The vertical line through the point (a, b) has equation x = a since every x-coordinate on the line has the same value a. The horizontal line through (a, b) has equation y = b since every y-coordinate on the line has the same value b. Finding Equations of Vertical and Horizontal Lines Horizontal Line is y = 8 Vertical Line is x = – 3 Y1 = 2x + 7 x y -3 1 0 7 (𝟎, 𝟕) −𝟔 (−𝟑, 𝟏) 1 7 6 m= 2 = 30 3 y – intercept ( 0 , 7 ) −𝟑 Slope y-intercept form y = mx + b slope y-intercept (0, b) General Linear Equation Although the general linear form helps in the quick identification of lines, the slope-intercept form is the one to enter into a calculator for graphing. Ax + By = C By = – Ax + C y = – (A/B) x + C/B Analyzing and Graphing a General Linear Equation Find the slope and y-intercept of the line 2𝑥 − 3𝑦 = 15 Rearrange for y −𝟑𝒚 = −𝟐𝒙 + 𝟏𝟓 −𝟑 −𝟐 𝟏𝟓 𝒚 = 𝒙+ −𝟑 −𝟑 −𝟑 𝟐 𝒚= 𝒙−𝟓 𝟑 Slope is 𝟐 𝟑 y-intercept is (𝟎, −𝟓) EXAMPLES State the slopes and y-intercepts of the given linear functions. 1. y = 4x 2. y = 3x – 5 1 3. 𝑓 𝑥 = 𝑥 − 2 3 4 slope = m = _______ 3 slope = m = _______ 𝟏 𝟑 slope = m = _______ y -intercept ( 0 , 0 ) y -intercept ( 0 , - 5 ) y -intercept ( 0 ,, - 2 ) Find the slope and y-intercept of the following 3 lines 4. 2𝑥 − 3𝑦 = 15 𝟐 𝟑 slope = m = _______ y-intercept 𝟏 𝟐 slope = m = _______ y-intercept 𝟎, 𝟑 𝟐 𝟓 𝟑 slope = m = _______ 𝟎, 𝟒 𝟑 𝟎, −𝟓 −𝟑𝒚 = −𝟐𝒙 + 𝟏𝟓 𝟐 𝒚= 𝒙−𝟓 𝟑 5. x + 2y = 3 𝟐𝒚 = −𝒙 + 𝟑 𝟏 𝟑 𝒚= 𝒙+ 𝟐 𝟐 6. 5x – 3y = – 4 −𝟑𝒚 = −𝟓𝒙 − 𝟒 𝟓 𝟒 𝒚= 𝒙+ 𝟑 𝟑 y-intercept Example 7 𝟐 Find the equation in slope-intercept form for the line with slope and 𝟑 passes through the point (−𝟑, 𝟓) Step 1: Solve for b using the point (−𝟑, 𝟓) 𝟐 𝟓 = −𝟑 + 𝒃 𝟑 b=7 Step 2: Find the equation 𝟐 𝟑 𝒚= 𝒙+𝟕 (𝟎, 𝟕) (−𝟑, 𝟓) Example 8 Find the equation in slope-intercept form for the line parallel to 𝒚 = and through the point (10, -1) Step 1: The slope of a parallel line will be 𝟐 𝒙 𝟓 + 𝟐 𝟐 𝟓 Step 2: Solve for b using the point (𝟏𝟎, −𝟏) 𝟐 −𝟏 = 𝟏𝟎 + 𝒃 𝟓 𝒃 = −𝟓 Step 3: Find the equation 𝒚 = 𝟐 𝒙 𝟓 (𝟏𝟎, −𝟏) –𝟓 (𝟎, −𝟓) Example 9 Write the equation for the line through the point (– 1 , 2) that is parallel to the line L: y = 3x – 4 Step 1: Slope of L is 3 so slope of any parallel line is also 3. Step 2: Find b. 𝟐 = 𝟑(−𝟏) + 𝒃 𝒃 = 𝟓 Step 3: The equation of the line parallel to L: 𝒚 = 𝟑𝒙 – 𝟒 is 𝒚 = 𝟑𝒙 + 𝟓 Step 4: Graph on your calculator to check your work. Use a square window. Y1 = 3x – 4 Y2 = 3x + 5 (0, 5) (0, – 4) Example 10 Write the equation for the line that is perpendicular to 𝟐 𝒚 = 𝒙 + 𝟐 and passes through the point (10, – 1 ) 𝟓 Step 1: The slope of a perpendicular line will be – 𝟓 𝟐 Step 2: Solve for b using the point (10, – 1) 𝟓 −𝟏 = – (𝟏𝟎) + 𝒃 Step 3: The equation of the line ┴ to 𝟐 𝟐 𝟓 𝒚 = 𝒙 + 𝟐 is 𝒚 = – 𝒙 + 𝟐𝟒 𝟓 𝟐 𝒃 = 𝟐𝟒 Step 4: Graph on your calculator to check your work. Use a square window. 𝟐 Y1 = 𝒙 + 𝟐 𝟓 5 2 Y2 = – x + 24 Example 11 Write the equation for the line through the point (– 1, 2) that is perpendicular to the line L: y = 3x – 4 Step 1: Slope of L is 3 so slope of any perpendicular line is − 𝟏 𝟑. Step 2: Find b. 𝟏 𝟐 = − (−𝟏) + 𝒃 𝟑 𝟓 𝒃 = 𝟑 Step 3: Find the equation of the line perpendicular to L: y = 3x – 4 𝟏 𝟓 𝒚 =− 𝒙 + 𝟑 𝟑 Step 4: Graph on your calculator to check your work. Use a square window. Y1 = 3x – 4 𝟏 𝟓 Y2 = − 𝒙 + 𝟑 𝟑 Example 12 Find the equation in slope-intercept form for the line that passes through the points (7, −2) and (−5, 8). Step 1: Find the slope Step 2: Solve for b using either point −𝟐 − 𝟖 −𝟏𝟎 𝟓 𝒎= = =− 𝟕 − (−𝟓) 𝟏𝟐 𝟔 𝟓 −𝟐 = − (𝟕) + 𝒃 𝟔 𝟐𝟑 𝒃 = 𝟔 𝟓 𝟖 = − (− 𝟓) + 𝒃 𝟔 𝒃 = Step 3: Find the equation 𝒚 = 𝟓 − 𝒙 𝟔 𝟐𝟑 + 𝟔 (– 4, 8) (7, – 2) 𝟐𝟑 𝟔 Example 13 Write the slope-intercept equation for the line through (– 2, –1) and (5, 4). −𝟏−𝟒 Slope = m = −𝟐−𝟓 = −𝟓 −𝟕 = 𝟓 𝟕 𝟓 – 𝟏 = (– 𝟐) + 𝒃 𝟕 𝒃 = 𝟑 𝟕 Equation for the line is 𝟓 𝟑 𝒚 = 𝒙 + 𝟕 𝟕 (5, 4) (– 2, – 1) UNIT 1B LESSON 3 SLOPE OF A SECANT 28 DEFINITION: A secant is a line that cuts a curve in two or more distinct points. P2 P1 P2 P3 P1 SLOPE OF A SECANT 29 EXAMPLE For the function y x 1 3 draw the curve and the secant line through the points A and C 2 X Y A -1 1 B 0 -2 C 1 -3 D 2 -2 E 3 1 A(-1,1) C(1, -3) Equation of AC 𝒎= −𝟑 − 𝟏 = −𝟐 𝟏 − (−𝟏) 𝟏 = −𝟐 −𝟏 + 𝒃 𝒃 = −𝟏 𝒚 = −𝟐𝒙 – 𝟏 𝒎<𝟎 30 EXAMPLE For the function y x 1 3 draw the curve and the secant line through the points B and D. 2 X Y A -1 1 B 0 -2 C 1 -3 D 2 -2 E 3 1 B(0,-2) D(2, -2) m=0 Equation of BD m 2 (2) 0 02 −𝟐 = 𝟎(𝟎) + 𝒃 𝒃 = −𝟐 𝒚 = −𝟐 31 EXAMPLE For the function y x 1 3 draw the curve and the secant lines through the points B and E. 𝒎>𝟎 2 X Y A -1 1 B 0 -2 C 1 -3 D 2 -2 E 3 1 E(3, 1) B(0,-2) Equation of BE −𝟐 − 𝟏 −𝟏 𝒎= = =𝟏 𝟎−𝟑 −𝟏 −𝟐 = 𝟏(𝟎) + 𝒃 𝒃 = −𝟐 y=x–2 Notice: The slopes of secants on curves change as the points change. Slopes may be positive, zero or negative. 32