Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 2 Graphs and Functions Copyright © 2013, 2009, 2005 Pearson Education, Inc.

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

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 3 Point-Slope Form The point–slope form of the equation of the line with slope m passing through the point (x 1, y 1 ) is

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 4 Example 1 USING THE POINT-SLOPE FORM (GIVEN A POINT AND THE SLOPE) Write an equation of the line through (– 4, 1) having slope – 3. Solution Point-slope form Be careful with signs. Distributive property Add 1.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 5 Example 2 USING THE POINT-SLOPE FORM (GIVEN TWO POINTS) Write an equation of the line through (– 3, 2) and (2, – 4). Write the result in standard form Ax + By = C. Solution Find the slope first. Definition of slope

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 6 Example 2 USING THE POINT-SLOPE FORM (GIVEN TWO POINTS) Solution Point-slope form x 1 = – 3, y 1 = 2, m = – 6/5 Multiply by 5. Write an equation of the line through (– 3, 2) and (2, – 4). Write the result in standard form Ax + By = C. Distributive property. Standard form

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 7 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 x 1 = 0 and y 1 = b gives the following. Slopey-intercept

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 8 Slope-Intercept Form The slope-intercept form of the equation of the line with slope m and y-intercept b is

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 9 Example 3 FINDING THE SLOPE AND y-INTERCEPT FROM AN EQUATION OF A LINE 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. mb The slope is and the y-intercept is –2.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 10 Example 4 USING THE SLOPE-INTERCEPT FORM (GIVEN TWO POINTS) Write 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. Substitute 3 for m in y = mx + b and choose one of the given points, say (1,1), to find b.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 11 Example 4 Solution Slope-intercept form m = 3, x = 1, y = 1 Solve for b. y-intercept USING THE SLOPE-INTERCEPT FORM (GIVEN TWO POINTS) The slope intercept form is Plot (0,–2) and then use the definition of slope to arrive at (1,1).

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 12 Example 5 FINDING AN EQUATION FROM A GRAPH Use the graph of the linear function  shown to complete the following. Solution The line falls 1 unit each time the x-value increases by 3 units. (a) Find the slope, y-intercept, and x-intercept. – 3– 3 y =  (x) – 1– 1

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 13 Example 5 FINDING AN EQUATION FROM A GRAPH Solution The graph intersects the y-axis at the point (0,– 1) and intersects the x-axis at the point (–3,0). The y-intercept is – 1 and the x-intercept is –3. – 3– 3 y =  (x) – 1– 1

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 14 Example 5 FINDING AN EQUATION FROM A GRAPH Use the graph of the linear function  shown to complete the following. (b) Write the equation that defines . – 3– 3 y =  (x) – 1– 1 Solution The slope is m = and the y-intercept is b = –1.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 15 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.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 16 Parallel Lines Two distinct nonvertical lines are parallel if and only if they have the same slope.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 17 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.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 18 Example 6 FINDING EQUATIONS OF PARALLEL AND PERPENDICULAR LINES Write the equation in both slope-intercept and standard form of the line that passes through the point (3, 5) and satisfies the given condition. Solution The point (3, 5) is on the line, so we need only to find the slope to use the point-slope form. We find the slope by writing the equation of the given line in slope- intercept form. (That is, we solve for y.) (a) parallel to the line 2x + 5y = 4

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 19 Example 6 FINDING EQUATIONS OF PARALLEL AND PERPENDICULAR LINES Solution (a) parallel to the line 2x + 5y = 4 Subtract 2x. Divide by 5. Write the equation in both slope-intercept and standard form of the line that passes through the point (3, 5) and satisfies the given condition.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 20 Example 6 FINDING EQUATIONS OF PARALLEL AND PERPENDICULAR LINES Solution (a) parallel to the line 2x + 5y = 4 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. Write the equation in both slope-intercept and standard form of the line that passes through the point (3, 5) and satisfies the given condition.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 21 Example 6 FINDING EQUATIONS OF PARALLEL AND PERPENDICULAR LINES Solution (a) parallel to the line 2x + 5y = 4 Point-slope form m = – 2/5, x 1 = 3, y 1 = 5 Distributive property Write the equation in both slope-intercept and standard form of the line that passes through the point (3, 5) and satisfies the given condition.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 22 Example 6 FINDING EQUATIONS OF PARALLEL AND PERPENDICULAR LINES Solution (a) parallel to the line 2x + 5y = 4 Add 5 = 25/5. Write the equation in both slope-intercept and standard form of the line that passes through the point (3, 5) and satisfies the given condition. Multiply by 5. Add 2x. slope-intercept form standard form

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 23 Example 6 FINDING EQUATIONS OF PARALLEL AND PERPENDICULAR LINES Solution In part (a) we found that the slope of the line 2x + 5y = 4 is –2/5. The slope of any line perpendicular to it is 5/2. (b) perpendicular to the line 2x + 5y = 4 Write the equation in both slope-intercept and standard form of the line that passes through the point (3, 5) and satisfies the given condition.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 24 Example 6 FINDING EQUATIONS OF PARALLEL AND PERPENDICULAR LINES Solution Distributive property Add 5 = 10/2. (b) perpendicular to the line 2x + 5y = 4 Write the equation in both slope-intercept and standard form of the line that passes through the point (3, 5) and satisfies the given condition. Multiply by 2. Subtract 2y, add 5, and rewrite. slope-intercept form standard form

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 25 EquationDescriptionWhen to Use y = mx +bSlope-Intercept Form Slope is m. y-intercept is b. Slope and y-intercept easily identified and used to quickly graph the equation. Also used to find the equation of a line given a point and the slope. y – y 1 = m(x – x 1 )Point-Slope Form Slope is m. Line passes through (x 1, y 1 ) Ideal for finding the equation of a line if the slope and a point on the line or two points on the line are known.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 26 EquationDescriptionWhen to Use Ax + By = CStandard Form (If the coefficients and constant are rational, then A, B, and C are expressed as relatively prime integers, with A ≥ 0). The x- and y- intercepts can be found quickly and used to graph the equation. The slope must be calculated.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 27 EquationDescriptionWhen to Use y = bHorizontal Line Slope is 0. y-intercept is b. If the graph intersects only the y-axis, then y is the only variable in the equation. x = aVertical Line Slope is undefined. x-intercept is a. If the graph intersects only the x-axis, then x is the only variable in the equation.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 28 Example 7 FINDING AN EQUATION OF A LINE THAT MODELS DATA Average annual tuition and fees for in-state students at public four-year colleges are shown in the table for selected years and graphed as ordered pairs of points where x = 0 represents 2005, x = 1 represents 2006, and so on, and y represents the cost in dollars. This graph of ordered pairs of data is called a scatter diagram.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 29 Example 7 FINDING AN EQUATION OF A LINE THAT MODELS DATA YearCost (in dollars)

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 30 Example 7 FINDING AN EQUATION OF A LINE THAT MODELS DATA (a) Find an equation that models the data. Solution The points lie approximately on a straight line. Write a linear equation that models the relationship between year x and cost y. Choose two data points, (0, 5492) and (5, 7605) to find the slope of the line.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 31 Example 7 FINDING AN EQUATION OF A LINE THAT MODELS DATA Solution (a) Find an equation that models the data.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 32 Example 7 FINDING AN EQUATION OF A LINE THAT MODELS DATA Solution The slope indicates that the cost of tuition and fees increased by about $423 per year. Use this slope, the y-intercept 5492, and the slope-intercept form to write the equation of the line. (a) Find an equation that models the data.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 33 Example 7 FINDING AN EQUATION OF A LINE THAT MODELS DATA Solution (a) Find an equation that models the data.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 34 Example 7 FINDING AN EQUATION OF A LINE THAT MODELS DATA Solution The value x = 7 corresponds to the year 2012, so we substitute 7 for x. (b) Use the equation from part (a) to predict the cost of tuition and fees at public 4-year colleges in 2012.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 35 Example 7 FINDING AN EQUATION OF A LINE THAT MODELS DATA Solution Model from part (a) Let x = 7. (b) Use the equation from part (a) to predict the cost of tuition and fees at public 4-year colleges in 2012.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 36 Example 7 FINDING AN EQUATION OF A LINE THAT MODELS DATA Solution The model predicts that average tuition and fees for in-state students at public four-year colleges in 2012 would be about $8450. (b) Use the equation from part (a) to predict the cost of tuition and fees at public 4-year colleges in 2012.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 37 Guidelines for Modeling 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.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 38 Example 8 SOLVING AN EQUATION WITH A GRAPHING CALCULATOR Solution Subtract 3x and 4. Use a graphing calculator to solve We write an equivalent equation with 0 on one side. Then we graph Y = –2X – 4(2 – X) – 3X – 4 to find the x-intercept. The standard viewing window cannot be used because the x-intercept does not lie in the interval [–10,10].

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 39 Example 8 SOLVING AN EQUATION WITH A GRAPHING CALCULATOR Solution Use a graphing calculator to solve As seen in the figure, the x-intercept of the graph is –12, and thus the solution of the equation is –12. (–12 is the zero of the function Y.) The solution set is {–12}.