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Section 2.2 - Slope and Rate of Change
ALGEBRA TWO CHAPTER TWO: LINEAR EQUATIONS AND FUNCTIONS Section Slope and Rate of Change
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LEARNING GOALS Goal One - Find slopes of lines and classify parallel and perpendicular lines. Goal Two - Use slope to solve real-life problems.
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VOCABULARY The slope of a nonvertical line is the ratio of vertical change (the rise) to the horizontal change (the run). The slope of a line is represented by the letter m. 1
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The Slope of a Line The slope of the nonvertical line passing through the points (x1, y1) and (x2, y2) is: slope=m = (y2-y1) (x2-x1) Rise: Difference of y values. Run: Difference of x values. 1
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Finding the slope of a Line
PROBLEM: Find the slope of a line passing through (-3, 2) and (1, 5). SOLUTION Let (x1, y1) = (-3, 2) and (x2, y2) = (1, 5) slope=m = (y2-y1) (x2-x1) Rise: Difference of y values. Run: Difference of x values. = (5 - 2)/(1-(-3)) Substitute values. = (3)/(1+3) = 3/4 Simplify. Slope is positive. The slope is 3/4. Because the slope is positive, the line rises from left to right.
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Finding the slope of a Line
PROBLEM: Find the slope of a line passing through (4, 1) and (-6, 3). SOLUTION Let (x1, y1) = (4, 1) and (x2, y2) = (-6, 3) slope =m= (y2-y1) (x2-x1) Rise: Difference of y values. Run: Difference of x values. = (3 - 1)/(-6-(4)) Substitute values. = (2)/(-10) = -1/5 Simplify. Slope is negative. The slope is -1/5. Because the slope is negative, the line falls from left to right.
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Parallel and Perpendicular Lines
PARALLEL LINES - The lines are parallel if and only if they have the same slope. Parallel lines NEVER intersect. m1 = m2 These two lines have the same slope, but different y-intercepts. They are parallel. 1
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Parallel and Perpendicular Lines
PERPENDICULAR LINES - The lines are perpendicular if and only if their slopes are negative reciprocals of each other. Perpendicular lines intersect to form a right angle. m1 = -1/m2 These two lines intersect at right angles and are perpendicular. Their slopes are m and -1/m . 1
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Parallel and Perpendicular Lines
PROBLEM: Tell whether the lines are parallel, perpendicular, or neither. Line 1: through (-1, 3) and (1, -3) Line 2: through (3, 0) and (0, 1) SOLUTION slope = (y2-y1) (x2-x1) Rise: Difference of y values. Run: Difference of x values. Line 1 Line 2 = (-3 -(-1)/(1-(-1)) = (1 - 0)/(0 - 3) = (-3 + 1)/(1 + 1) = (1)/(-3) = (-2)/(2) = -1 = -1/3 1
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Line 1: through (-1, 3) and (1, -3) Line 2: through (3, 0) and (0, 1)
Parallel and Perpendicular Lines PROBLEM: Tell whether the lines are parallel, perpendicular, or neither. Line 1: through (-1, 3) and (1, -3) Line 2: through (3, 0) and (0, 1) SOLUTION These lines are neither parallel or perpendicular. Line 1 Line 2 = (-3 -(-1)/(1-(-1)) = (1 - 0)/(0 - 3) = (-3 + 1)/(1 + 1) = (1)/(-3) = (-2)/(2) = -1 = -1/3 1
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Parallel and Perpendicular Lines
PROBLEM: Tell whether the lines are parallel, perpendicular, or neither. Line 1: through (-2, 6) and (0, -2) Line 2: through (-1, 6) and (1, -2) SOLUTION slope = (y2-y1) (x2-x1) Rise: Difference of y values. Run: Difference of x values. Line 1 Line 2 = (-2 - 6)/(0 -(-2)) = (-2 - 6)/(1 - (-1)) = (-8)/(1 + 1) = (-8)/(2) = (-8)/(2) = -4 = -4 1
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Parallel and Perpendicular Lines These lines are parallel.
PROBLEM: Tell whether the lines are parallel, perpendicular, or neither. Line 1: through (-2, 6) and (0, -2) Line 2: through (-1, 6) and (1, -2) SOLUTION These lines are parallel. Line 1 Line 2 = (-2 - 6)/(0 -(-2)) = (-2 - 6)/(1 - (-1)) = (-8)/(1 + 1) = (-8)/(2) = (-8)/(2) = -4 = -4 1
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Average rate of change = (change in value) / (change in time)
Slope as Rate of Change PROBLEM: Six years ago a house was purchased for $89,000. This year it was appraised at $125,000. Find the average rate of change and use it to determine the value after nine years. SOLUTION Average rate of change = (change in value) / (change in time) = ($125,000 - $89,000)/6 years = $36,000/6 years = $6,000 per year Over a 9-year period, the value changed 9 • $6, 000/yr. The value after nine years was $89, $54,000 = $143,000.
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Homework Pg 79-80 #17,23,27,29,37,39,41,43,51
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