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Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Chapter 3 Introduction to Graphing.

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Presentation on theme: "Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Chapter 3 Introduction to Graphing."— Presentation transcript:

1 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Chapter 3 Introduction to Graphing

2 3-2 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Point-Slope Form and Equations of Lines Point-Slope Form Graphing the Equation of a Line Estimations and Predictions Using Two Points 3.7

3 3-3 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Graph Solution Find an equivalent equation: The line passes through (  2,  3) and has slope  4/3. down 4 right 3 (  2,  3)

4 2-4 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Suppose that a line of slope m passes through the point (x 1, y 1 ). For any other point (x, y) to lie on this line, we must have (x 1, y 1 ) (x, y) x – x 1 y – y 1 Point-Slope Form

5 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. 2-5 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Point-Slope Form Any equation y – y 1 = m(x – x 1 ) is said to be written in point-slope form and has a graph that is a straight line. The slope of the line is m. The line passes through (x 1, y 1 ).

6 3-6 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Write a point-slope equation for the line with slope 2/3 that contains the point (4, 9). Solution We substitute 2/3 for m, and 4 for x 1, and 9 for y 1 :

7 3-7 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Write the slope-intercept equation for the line with slope 3 and point (4, 3). Solution There are two parts to this solution. First, we write an equation in point-slope form:

8 3-8 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Next, we find an equivalent equation of the form y = mx + b: Using the distributive law Adding 3 to both sides to get the slope-intercept form

9 3-9 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Find an equation for the line parallel to 5y = 3x + 10 with y-intercept (0, 6). Solution Solve for y to find the slope-intercept form: The slope of any parallel line will be 3/5. The slope-intercept form yields: The slope is 3/5.

10 3-10 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Find an equation for the line perpendicular to 5y = 3x + 10 that passes through (2, –4). Solution Solve for y to find the slope-intercept form: The slope of any perpendicular will be the opposite of the reciprocal or –5/3. The slope is 3/5.

11 3-11 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Use slope-intercept form to find an equation of the line with slope 2 that passes through (2, 6). Solution Since the slope of the line is 2, we have… To find b, substitute into the equation and solve. The equation of the line is: y = 2x + 2.

12 3-12 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Find the equation of the horizontal line that passes through (3, –2). Solution The equation of a horizontal line is of the form y = b. In order for (3, –2) to be a solution of y = b, we must have b = –2. The equation of the line is y = –2.

13 3-13 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Find the equation of the vertical line that passes through (3, –2). Solution The equation of a vertical line is of the form x = a. In order for (3, –2) to be a solution of x = a, we must have a = 3. The equation of the line is x = 3.

14 3-14 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Estimations and Predictions Using Two Points It is possible to use line graphs to estimate real-life quantities that are not already known. To do so, we calculate the coordinates of an unknown point by using two points with known coordinates. When the unknown point is located between the two points, this process is called interpolation. Sometimes a graph passing through the known points is extended to predict future values. Making predictions in this manner is called extrapolation.

15 3-15 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Aerobic exercise. A person’s target heart rate is the number of beats per minute that bring the most aerobic benefit to his or her heart. The target heart rate for a 20-year-old is 150 beats per minute and for a 60-year-old, 120 beats per minute. a) Graph the given data and calculate the target heart rate for a 46-year-old. b) Calculate the target heart rate for a 70-year-old.

16 3-16 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example a) We draw the axes and label, using a scale that will permit us to view both the given and the desired data. The given information allows us to then plot (20, 150) and (60, 120).

17 3-17 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example We determine the slope of the line. Use one point and write the equation of the line.

18 3-18 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example a) To calculate the target heart rate for a 46-year-old, we substitute 46 for x in the slope-intercept equation: The graph confirms the target heart rate.

19 3-19 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example b) To calculate the target heart rate for a 70-year-old, we substitute 70 for x in the slope-intercept equation: The graph confirms the target heart rate.


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