Chapter 1 Functions and Their Graphs

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Presentation transcript:

Chapter 1 Functions and Their Graphs PreCalculus Chapter 1 Functions and Their Graphs

Chapter 1 - Contents 1-1 Lines in the Plane 1-2 Functions 1-3 Graphs of Functions 1-4 Shifting, Reflecting, and Stretching Graphs 1-5 Combinations of Functions 1-6 Inverse Functions 1-7 Linear Models and Scatter Plots

Chapter 1 - Objectives Write linear equations. Graph linear equations. Evaluate functions. Find the domain and range of a function. Graph functions. Graph function transformations. Use functions in modeling real-world data.

Lines in the Plane Section 1-1

Objectives Find the slopes of lines. Write linear equations given points on lines and their slopes. Use slope intercept form of linear equations to sketch lines. Use slope to identify parallel and perpendicular lines.

Linear Equations Equations of the form ax + by = c are called linear equations in two variables. x y 2 -2 This is the graph of the equation 2x + 3y = 12. (0,4) (6,0) The point (0,4) is the y-intercept. The point (6,0) is the x-intercept.

Slope of a Line y x 2 -2 m = 0 m = 2 m is undefined m = 1 m = - 4 The slope of a line is a number, m, which measures its steepness.

Slope Formula The slope of the line passing through the two points (x1, y1) and (x2, y2) is given by the formula y2 – y1 x2 – x1 m = , (x1 ≠ x2 ). x y (x1, y1) (x2, y2) x2 – x1 y2 – y1 change in y change in x The slope is the change in y divided by the change in x as we move along the line from (x1, y1) to (x2, y2).

Study Tip Whenever using a formula (like slope formula or any other formula) in solving a problem, write the formula down first and then substitute in the correct values. Then solve the resulting equation for the known value.

Example: Find the slope of the line passing through the points (2, 3) and (4, 5). Use the slope formula with x1= 2, y1 = 3, x2 = 4, and y2 = 5. y2 – y1 x2 – x1 m = 5 – 3 4 – 2 = = 2 = 1 x y (4, 5) (2, 3) 2 2

Your Turn Find the slope of the line passing through each pair of points. (-5, -6) and (2, 8) (4, 5) and (8, 4)

Slope of Horizontal and Vertical Lines Slope of a horizontal line is 0 Equation of a horizontal line that passes through the point (a,b):

Slope of Horizontal and Vertical Lines Slope of a vertical line is undefined Equation of a vertical line that passes through the point (a,b):

y x 2 -2 m = 0 m is undefined Slope of a Line A line with positive slope (m>0) rises from left to right. A line with negative slope (m<0) falls from left to right. A line with zero slope (m=0) is horizontal. A line with undefined slope is vertical.

Point-Slope Form A linear equation written in the form y – y1 = m(x – x1) is in point-slope form. The graph of this equation is a line with slope m passing through the point (x1, y1). x y 4 8 Example: The graph of the equation y – 3 = - (x – 4) is a line of slope m = - passing through the point (4, 3). 1 2 m = - 1 2 (4, 3)

Point-Slope Form Useful for finding the equation of a line if you know at least one point the line passes through and the slope of the line. Derived from the from the slope formula. Slope Formula Change y2, x2 to just y and x. Multiple both sides by the denominator. Point-Slope Form

Example: Write the slope-intercept form for the equation of the line through the point (-2, 5) with a slope of 3. Use the point-slope form, y – y1 = m(x – x1), with m = 3 and (x1, y1) = (-2, 5). y – y1 = m(x – x1) Point-slope form y – y1 = 3(x – x1) Let m = 3. y – 5 = 3(x – (-2)) Let (x1, y1) = (-2, 5). y – 5 = 3(x + 2) Simplify. y = 3x + 11 Slope-intercept form

Your Turn Write the slope-intercept form for the equation of the line through the points (4, 3) and (-2, 5). 2 1 5 – 3 -2 – 4 = - 6 3 Calculate the slope. m = y – y1 = m(x – x1) Point-slope form Use m = - and the point (4, 3). y – 3 = - (x – 4) 1 3 Slope-intercept form y = - x + 13 3 1

Slope-Intercept Form A linear equation written in the form y = mx + b is in slope-intercept form. The slope is m and the y-intercept is (0, b). To graph an equation in slope-intercept form: 1. Write the equation in the form y = mx + b. Identify m and b. 2. Plot the y-intercept (0, b). 3. Starting at the y-intercept, find another point on the line using the slope. 4. Draw the line through (0, b) and the point located using the slope.

Example: Graph the line y = 2x – 4. The equation y = 2x – 4 is in the slope-intercept form. So, m = 2 and b = - 4. x y 2. Plot the y-intercept, (0, - 4). 1 = change in y change in x m = 2 3. The slope is 2. (1, -2) 2 4. Start at the point (0, 4). Count 1 unit to the right and 2 units up to locate a second point on the line. (0, - 4) 1 The point (1, -2) is also on the line. 5. Draw the line through (0, 4) and (1, -2).

Matching Examples Solution: A. B. C. 1) C 2) A 3) B

Example: Find the slope and y-intercept of the linear equation 2x+3y-4=0. Solve by writing the equation in slope-intercept form (ie. solve for y). Given Add 4 to each side Subtract 2x from each side Divide both sides by 3

Your Turn Determine the slope and y-intercept of the linear equation -4x-y+5=0. Answer: m=-4 and y-intercept is (0, 5)

Summary of Equations of Lines General form: Vertical line: Horizontal line: Slope-intercept form: Point-slope form:

Linear Model Writing an equation of a line that models real data: If the data changes at a fairly constant rate, the rate of change is the slope. An initial condition would be the y-intercept. Example: Suppose there is a flat rate of $.20 plus a charge of $.10/minute to make a phone call. Write an equation that gives the cost y for a call of x minutes. Note: The initial condition is the flat rate of $.20 and the rate of change is $.10/minute. Solution: y = .10x + .20

Linear Model Writing an equation of a line that models real data: If the data changes at a fairly constant rate, the rate of change is the slope. An initial condition would be the y-intercept. Example: The percentage of mothers of children under 1 year old who participated in the US labor force is shown in the table. Find an equation that models the data. Using (1980,38) and (1998,59)

Your Turn The net sales for a car manufacturer were $14.61 billion in 2005 and $15.78 billion in 2006. Write a linear equation giving the net sales y in terms of x, where x is the number of years since 2000. Then use the equation to predict the net sales for 2007. Answer: y=1.17x+8.76, predicted sales for 2007 is $16.95 billion.

Study Tip The prediction method illustrated in the last example is called extrapolation. Note in the top figure that an extrapolated point does not lie between the given points. When the estimated point lies between two given points, as shown in the bottom figure, the procedure used to predict the point is called linear interpolation.

Parallel Lines Two lines are parallel if they have the same slope. x y If the lines have slopes m1 and m2, then the lines are parallel whenever m1 = m2. y = 2x + 4 (0, 4) Example: The lines y = 2x – 3 and y = 2x + 4 have slopes m1 = 2 and m2 = 2. y = 2x – 3 (0, -3) The lines are parallel.

Perpendicular Lines Two lines are perpendicular if their slopes are negative reciprocals of each other. If two lines have slopes m1 and m2, then the lines are perpendicular whenever x y y = 3x – 1 1 m1 m2= - (0, 4) y = - x + 4 1 3 or m1m2 = -1. Example: The lines y = 3x – 1 and y = - x + 4 have slopes m1 = 3 and m2 = - . 1 3 (0, -1) The lines are perpendicular.

Finding equations of Parallel or Perpendicular lines: Example Finding equations of Parallel or Perpendicular lines: If parallel lines are required, the slopes are identical. If perpendicular lines are required, use slopes that are negative reciprocals of each other. Example: Find an equation of a line passing through the point (-8,3) and parallel to 2x - 3y = 10. Step 1: Find the slope Step 2: Use the point-slope method of the given line

Finding equations of Parallel or Perpendicular lines: Your Turn Finding equations of Parallel or Perpendicular lines: Find an equation of a line passing through the point (-8,3) and perpendicular to 2x - 3y = 10. Step 1: Find the slope Step 3: Use the point-slope method of the given line Step 2: Take the negative reciprocal of the slope found

Summary of Linear Graphs

Additional Problems Find the slope of the line passing through the points (-5, -6) and (2, 8). 2 Find an equation of the line that passes through the points (3, -7) and has slope of 2. y = 2x – 13 Determine the slope and y-intercept of the linear equation -4x – y + 5 = 0. Then graph. m = -4, y-int. (0, 5) Find the slope-intercept form of the equation of the line that passes through the point (-4, 1) and is parallel to the line 5x – 3y = 8. y = 5/3x + 23/3 Find the slope-intercept form of the equation of the line that passes through the point (-4, 1) and is perpendicular to the line 5x – 3y = 8. y = -3/5x – 7/5

Additional Problems A real estate office handles an apartment complex with 50 units. When the rent per unit is $580 per month, all 50 units are occupied. However, when the rent is $625 per month, the average number of occupied units drops to 47. assume that the relationship between the monthly rent p and the demand x is linear. Write the equation of the line giving the demand x in terms of the rent p. Use the demand equation to predict the number of units occupied when the rent is lowered to $595. Use the demand equation to predict the number of units occupied when the rent is $655. x = -p/15 + 266/3 49 units 45 units

Homework Section 1.1, pg. 11 – 15: Vocabulary Check #1 – 5 all Exercises: #11-49 odd, 53-61 odd Read Section 1.2, pg. 16-23