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4 th EDITION

College Algebra & Trigonometry and Precalculus. 4 th EDITION. 2.5. Equations of Lines; Curve Fitting. Point-Slope Form Slope-Intercept Form Vertical and Horizontal Lines Parallel and Perpendicular Lines Modeling Data Solving Linear Equations in One Variable by Graphing.

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4 th EDITION

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  1. College Algebra & Trigonometry and Precalculus 4th EDITION

  2. 2.5 Equations of Lines; Curve Fitting Point-Slope Form Slope-Intercept Form Vertical and Horizontal Lines Parallel and Perpendicular Lines Modeling Data Solving Linear Equations in One Variable by Graphing

  3. Point-Slope Form The line with slope m passing through the point (x1, y1) has an equation the point –slope form of the equation of a line.

  4. USING THE POINT-SLOPE FORM (GIVEN A POINT AND THE SLOPE) Example 1 Find an equation of the line through (–4, 1) having slope –3. Solution Point-slope form Be careful with signs. Distributive property Add 1.

  5. USING THE POINT-SLOPE FORM (GIVEN TWO POINTS) Example 2 Find an equation of the line through (–3, 2) and (2, –4). Solution Find the slope first. Definition of slope

  6. USING THE POINT-SLOPE FORM (GIVEN TWO POINTS) Example 2 Find an equation of the line through (–3, 2) and (2, –4). Solution Point-slope form x1 = –3, y1 = 2, m = –6/5 Multiply by 5.

  7. USING THE POINT-SLOPE FORM (GIVEN TWO POINTS) Example 2 Find an equation of the line through (–3, 2) and (2, –4). Solution Multiply by 5. Distributive property. Add 10; divide by 5.

  8. Slope-Intercept Form As a special case,suppose that a line passes through the point (0, b), so the line has y-intercept b. If the line has slope m, then using the point-slope form with x1 = 0 and y1 = b gives Slope y-intercept

  9. Slope-Intercept Form The line with slope m and y-intercept b has an equation the slope-intercept form of the equation of a line.

  10. FINDING THE SLOPE-INTERCEPT FORM (GIVEN TWO POINTS) Example 3 Find the slope and y-intercept of the line with equation 4x +5y = –10. Solution Write the equation in slope-intercept form. Subtract 4x. Divide by 5. m b

  11. USING THE SLOPE-INTERCEPT FORM (GIVEN TWO POINTS) Example 4 Find an equation of a line through (1, 1) and (2,4). Then graph the line using the slope-intercept form. Solution Use the slope intercept form. First, find the slope. Definition of slope.

  12. USING THE SLOPE-INTERCEPT FORM (GIVEN TWO POINTS) Example 4 Find an equation of a line through (1, 1) and (2,4). Then graph the line using the slope-intercept form. Solution Substitute 3 for m in y = mx + b and choose one of the given points, say (1, 1), to find b. Slope-intercept form m = 3, x = 1, y = 1 Solve for b. y-intercept

  13. USING THE SLOPE-INTERCEPT FORM (GIVEN TWO POINTS) Example 4 y Solution (2, 4) The slope intercept form is (1, 2) x (0, –2) y changes 3 units x changes 1 unit

  14. FINDING AN EQUATION FROM A GRAPH Example 5 Use the graph of the linear function  shown here to complete the following. a. Find the slope, y-intercept, and x-intercept. –3 Solution The line falls 1 unit each time the x-value increases by 3. –1 y = (x)

  15. FINDING AN EQUATION FROM A GRAPH Example 5 Use the graph of the linear function  shown here to complete the following. a. Find the slope, y-intercept, and x-intercept. –3 Solution The line intersects the y-axis at point (0, –1). The y-intercept is –1. –1 y = (x)

  16. FINDING AN EQUATION FROM A GRAPH Example 5 Use the graph of the linear function  shown here to complete the following. a. Find the slope, y-intercept, and x-intercept. –3 Solution The line intersects the x-axis at point ( –3, 0). The x-intercept is –3. –1 y = (x)

  17. FINDING AN EQUATION FROM A GRAPH Example 5 Use the graph of the linear function  shown here to complete the following. b. Write the equation that defines . –3 –1 y = (x)

  18. Equations of Vertical and Horizontal lines An equation of the vertical line through the point (a, b) is x = a. An equation of the horizontal line through the point (a, b) is y = b.

  19. Parallel Lines Two distinct nonvertical lines are parallel if and only if they have the same slope.

  20. Perpendicular Lines Two lines neither of which is vertical, are perpendicular if and only if their slopes have a product of –1. Thus, the slopes of perpendicular lines, neither of which are vertical, are negative reciprocals.

  21. FINDING EQUATIONS OF PARALLEL AND PERPENDICULAR LINES Example 6 Find the equation in slope-intercept form of the line that passes through the point (3, 5) and satisfies the given condition. a. parallel to the line 2x + 5y = 4 Solution (3, 5) is on the line so we need to find the slope to use the point-slope form. Write the equation in the slope-intercept form (solve for y).

  22. FINDING EQUATIONS OF PARALLEL AND PERPENDICULAR LINES Example 6 Find the equation in slope-intercept form of the line that passes through the point (3, 5) and satisfies the given condition. a. parallel to the line 2x + 5y = 4 Solution Subtract 2x. Divide by 5.

  23. FINDING EQUATIONS OF PARALLEL AND PERPENDICULAR LINES Example 6 Find the equation in slope-intercept form of the line that passes through the point (3, 5) and satisfies the given condition. a. parallel to the line 2x + 5y = 4 Solution The slope is – 2/5. Since the lines are parallel, – 2/5 is also the slope of the line whose equation is to be found.

  24. FINDING EQUATIONS OF PARALLEL AND PERPENDICULAR LINES Example 6 Find the equation in slope-intercept form of the line that passes through the point (3, 5) and satisfies the given condition. a. parallel to the line 2x + 5y = 4 Solution Point-slope form m = –2/5, x1 = 3, y1 = 5 Distributive property

  25. FINDING EQUATIONS OF PARALLEL AND PERPENDICULAR LINES Example 6 Find the equation in slope-intercept form of the line that passes through the point (3, 5) and satisfies the given condition. a. parallel to the line 2x + 5y = 4 Solution Distributive property Add 5 (25/5).

  26. FINDING EQUATIONS OF PARALLEL AND PERPENDICULAR LINES Example 6 Find the equation in slope-intercept form of the line that passes through the point (3, 5) and satisfies the given condition. b. perpendicular to the line 2x + 5y = 4 Solution We know the slope of the line, so the slope of any line perpendicular to it is 5/2.

  27. FINDING EQUATIONS OF PARALLEL AND PERPENDICULAR LINES Example 6 b. perpendicular to the line 2x + 5y = 4 Solution Distributive property Add 5 (10/2).

  28. FINDING AN EQUATION OF A LINE THAT MODELS DATA Example 7 y 6000 5000 4000 3000 x 2 4 6 0 8 10 Year

  29. FINDING AN EQUATION OF A LINE THAT MODELS DATA Example 7 y a. Find an equation that models the data. 6000 5000 Solution The points lie approximately on a straight line. Write an equation that models the relationship between year x and cost y. Use two data points (0, 3151) and (10, 5836). 4000 3000 x 2 4 6 0 8 10 Year

  30. FINDING AN EQUATION OF A LINE THAT MODELS DATA Example 7 y a. Find an equation that models the data. 6000 5000 Solution 4000 3000 x 2 4 6 0 8 10 Year

  31. FINDING AN EQUATION OF A LINE THAT MODELS DATA Example 7 y a. Find an equation that models the data. 6000 5000 Solution The slope 268.5 indicates that the cost of tuition and fees increased by about $269 per year. Use this slope and the y-intercept 3151 to write the equation of the line. 4000 3000 x 2 4 6 0 8 10 Year

  32. FINDING AN EQUATION OF A LINE THAT MODELS DATA Example 7 y a. Find an equation that models the data. 6000 5000 Solution 4000 3000 Slope-intercept form x 2 4 6 0 8 10 Year

  33. FINDING AN EQUATION OF A LINE THAT MODELS DATA Example 7 y b. Use the equation developed to predict the cost of tuition and fees at public 4-year colleges in 2008. 6000 5000 4000 Solution The value x = 12 corresponds to the year 2008, so we substitute 12 for x. 3000 x 2 4 6 0 8 10 Year

  34. FINDING AN EQUATION OF A LINE THAT MODELS DATA Example 7 y b. Use the equation developed to predict the cost of tuition and fees at public 4-year colleges in 2008. 6000 5000 4000 Solution 3000 Model from part (a). x 2 4 6 0 8 10 Let x = 12. Year

  35. FINDING AN EQUATION OF A LINE THAT MODELS DATA Example 7 y b. Use the equation developed to predict the cost of tuition and fees at public 4-year colleges in 2008. 6000 5000 4000 Solution According to the model, average tuition and fees in 2008 would be about $6373. 3000 x 2 4 6 0 8 10 Year

  36. Guidelines for Fitting a Curve Step 1 Make a scatter diagram of the data. Step 2 Find an equation that models the data. For a line, this involves selecting two data points and finding the equation of the line through them

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