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2-3 Slope. Slope indicates the steepness of a line. is “the change in y over the change in x” (vertical over horizontal). is the “ m” in y = m x + b. Finding the Slope of a Line. m = m = We use the first equation when given a graph, the second when given two points. x. x.
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2-3 Slope Slope • indicates the steepness of a line. • is “the change in y over the change in x” (vertical over horizontal). • is the “m” in y = mx + b.
Finding the Slope of a Line • m = • m = • We use the first equation when given a graph, the second when given two points.
x x y y Slope • If a line rises from left to right (uphill), it has a positive slope. • If a line falls from left to right (downhill), it has a negative slope.
x y Slope • If a line is vertical, it has an undefined slope. (Remember VUX – vertical/undefined/x = #).
x y Slope • If a line is horizontal it has a slope equal to 0. (Remember HOY – horizontal/zero/y = #).
y Find the Slope • To find the slope of this line we will use rise/run. • Pick either point. • We find the rise by counting up (+) or down (-). • Rise is 3 up from the bottom or -3 down from the top point. x
x y Find the Slope • To find the run, we count right (+) or left (-). • If you started from the top point, your run will be left (-2), if you started at the bottom point, your run will be right (2).
x y Find the Slope • So, the slope is either 3/2 or -3/-2, depending upon which point you started counting from. Either way, the slope is written as 3/2.
x x y y Find the Slope • m = -2/3 • m = -7/2
x x y y Find the Slope • m= undefined • m = zero
Find the Slope of the Line • Given the points (5, 1) & (6, 4) m = = = 3
Parallel Lines • Parallel lines have the SAME SLOPE! • What is the slope of the line parallel to y = -2x + 4? • m = -2
Perpendicular Lines • Perpendicular lines have slopes that are opposite reciprocals! • Opposite reciprocals multiply with each other to get an answer of -1. • What is the slope of the line perpendicular to y = -2x + 4? • m = ½ • ½ is the opposite reciprocal of -2. • -2(1/2) = -1
x y Graphing a Line Given a Point & Slope • Graph a line though the point (2, -6) with m=2/3 • Graph (2, -6) • Count up 2 for the rise, and to the right 3 for the run • Plot the new point, then connect
Graphing Lines • Graph the line perpendicular to y = 2x + 3 that goes through the point (-2, 3). • The slope of the line is 2 so the slope of the perpendicular line is -1/2. • 2(-1/2) = -1 • m = -1/2
x y Graphing Lines • m = - ½ • Plot point given (-2,3) • Use the slope to find two more points • Connect
x y Graphing Lines • Check by graphing original line to make sure they at least look perpendicular. The original line was y = 2x + 3 That means that it crosses y at 3, with a slope of 2. Graphing that line gives us visual confirmation that we are at least close to being right with our perpendicular line.
Graphing Lines • Graph the line parallel to x = -1 that goes through the point (3, -3). • The slope of the line is undefined (VUX) so the slope of the parallel line is undefined.
x y Graph the Line • (3, -3) m = undefined
Slope-Intercept Form • y = mx + b [ f(x) = mx + b] • m is the slope • b is the y-intercept
x y Find the Equation of the Line • What is the y-intercept? • b = -2 • What is m? • m = • y = x -2
x y Find the Equation of the Line • b = ? • b = 1 • m = ? • m = - • y = -x + 1
Find the Slope and y-intercept of each Equation: 1) y = -4x + 3 m = -4 b = 3 2) y = 5 - x Rewrite this to y = - x + 5 m = - b = 5
Find the Equation Given the Slope and y-intercept • m = -3, b = 1 y = -3x + 1 2) m = -, b = - 4 y = - x – 4
Find the Equation Given the Slope and a Point Given m = -1, (2, 1) First, calculate b by substituting the slope and the coordinates into y = mx + b. y = -1x + b 1 = -1(2) + b 1 = -2 + b 3 = b y = -x + 3
Find the Equation Given Two Points Given (-1, 3) and (2, 1) First, calculate the slope: Second, find b. Use either point... m = , (2, 1) 1 = (2) + b 1 = + b = b
Point-Slope Form of an Equation (y – y1) = m(x – x1) Example: Write the Point-Slope equation of the line with a slope of 3 that goes through the point (6, -4). (y – -4) = 3(x – 6) (y + 4) = 3(x – 6)
Standard Form of an Equation Ax + By = C, A and B cannot both be equal to zero, and should be expressed as integers. Basically, solve for the constant (C). Example: 4y = 9x – 2 4y – 9x = -2 -9x + 4y = -2
Standard Form of an Equation Ax + By = C, A and B cannot both be equal to zero, and should be expressed as integers. Example: 2/3x + 8 = 3/5y 8 = -2/3x + 3/5y To express A and B as integers, we must find a common denominator and multiply both sides by that number.
Standard Form of an Equation Ax + By = C, A and B cannot both be equal to zero, and should be expressed as integers. Example: 2/3x + 8 = 3/5y 8 = -2/3x + 3/5y 15 is the LCD of 3 and 5. (-2/3x)(15) = -10x (3/5y)(15) = 9y 8(15) = 120 -10x + 9y = 120
Review Question Find the equations of the line that goes through the point (4,-6) and is perpendicular to y = 4x – 1 in slope-intercept, point-slope, and standard forms. Need to know the slope for all three forms. Slope of a perpendicular line is the opposite reciprocal of the original slope. Opposite reciprocal of 4 is -1/4
Review Question Slope-Intercept Form y = mx + b (y + 6) = -1/4(x - 4) y + 6 = -1/4x + 1 y = -1/4x - 5 Point-Slope Form (y - y1) = m(x - x1) (4, -6), slope of -1/4 (y - -6) = -1/4(x – 4) (y + 6) = -1/4(x - 4)
Review Question Standard Form Ax + By = C (A & B are integers) Let’s take the slope-intercept form and solve to get constant by itself. y = -1/4x – 5 1/4x + y = -5 Multiply by 4 to make A an integer. x + 4y = -20
Assignments: • Classwork: • Pg. 79 #38-60 Even • Pgs. 86-87 #30-56 Even • Homework: • Pg. 78 # 8-36 Even • Pg. 86 # 10-28 Even