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### Chapter One

Functions and Their Graphs

1.1 – Lines in the Plane

- Slope
- Equation of a line
- Point-Slope form
- Slope-intercept form
- General form
- Parallel and Perpendicular Lines
- Applications

Slope

- Formula: (y2-y1)/ (x2-x1)
- Denoted: m
- In words: Change in y over change in x

Slope Generalizations

- Positive slope – rises (left to right)
- Negative slope – falls (left to right)
- Horizontal line – slope = 0
- Vertical line – slope = undefined

Example 1.1.1

Pg. 11 # 7 (Just find the slope)

Find the slope of the line from the given points.

Points = (0,-10) and (-4,0)

Line Equations

- Point-Slope form
- Y-intercept form
- General form

Y-Intercept Form

- Most common
- Formula: y = mx + b
- Can be found by solving for y in the point-slope form or the general form.
- Easy to sketch a line from this equation.

General Form

- Formula: Ax + By = C
- Can also be found from either of the other forms.

Example 2.1.1

Pg. 12 # 25

Find the general form of the line from the given point and slope.

Point = (0,-2) Slope = 3

Example 3.1.1

Pg. 12 # 29

Find the general form of the line from the given point and slope.

Point = (6,-1) Slope = undefined

Solution - Ex. 3.1.1

If the slope is undefined we know that it is a vertical line. A vertical line only crosses the x-axis. So all we need is the x-value from the point given.

x = 6

This is also in general form.

Parallel and Perpendicular Lines

- If two lines are parallel they have the same slope.
- If two lines are perpendicular their slopes are negative reciprocals of each other.
- “Change the sign and flip it!”

Example 4.1.1

Pg. 12 # 57

Write the slope-intercept form of the equation of the line through the give point

a) parallel to the line given.

b) perpendicular to the line given.

Point = (2,1) Line => 4x -2y = 3

Solution - Ex. 4.1.1

Solving the given equation for y we get:

y = 2x – 3/2

- same slope (use the point-slope form)

(y – 1) = 2(x – 2)

y – 1 = 2x – 4

y = 2x – 3

Solution - Ex. 4.1.1

b) negative reciprocal slope (use the point-slope form).

(y – 1) = (-1/2)(x – 2)

y – 1 =(-1/2x) + 1

y = (-1/2x) + 2

Example 5.1.1

Pg. 12 # 53

Determine whether the lines L1 and L2 passing through the pairs of points are parallel, perpendicular, or neither.

L1: ( 0,-1) and (5,9)

L2: (0, 3) and (4,1)

Solution - Ex. 5.1.1

L1: Slope = (9 + 1)/(5 – 0) = 2

L2: Slope = (1 – 3)/(4 – 0) = -2/4 = -1/2

m1 * m2 = -1 so these are negative reciprocals

L1 and L2 are perpendicular.

Example 6.1.1

Pg. 13 # 71

Looking at the graph:

a) Use the slopes to determine the year(s) in which the earnings per share of stock showed the greatest increase and decrease.

b) Find the equation of the line between the years 1992 and 2002.

Example 6.1.1

Pg. 13 # 71 (cont.)

c) Interpret the meaning of the slope from part (b) in the context of the problem.

d) Use the equation from part (b) to estimate the earnings per share of stock for the year 2006. Do you think this is an accurate estimation? Explain.

Solution - Ex. 6.1.1

a) The greatest increase was between 1998 and 1999.

b) 2 points (2, 0.58) and (12, 0.08) gives the slope -0.05.

Use point-slope: (y – 0.58)=-0.05(x-2)

y =- 0.05x + 0.68

Solution - Ex. 6.1.1

c) For every year increase there is a 0.05 decrease in earning per share.

d) Plug in the appropriate number:

y= -0.05(16) + 0.68

= -0.12

This is not accurate because our line does not accurately represent the data.

Example 7.1.1

Pg. 14 # 85

A controller purchases a bulldozer for $36,500. The bulldozer requires an average expenditure of $5.25/hr for fuel and maintenance, and the operator is paid $11.50/hr.

a) Write a linear equation giving the total cost (C) of operating the bulldozer for t hours. (Include the purchase cost of the bulldozer)

Example 7.1.1

Pg. 14 # 85 (cont.)

b) Assuming that customers are charged $27/hr of bulldozer use, write an equation for the revenue (R) derived from t hours of use.

c) Use the profit formula (P=R-C) to write an equation for the profit derived from t hours of use.

Example 7.1.1

Pg. 14 # 85 (cont.)

d) Use the result of part (c) to find the break-even point. (The number of hours the bulldozer must be used to yield a profit of $0.

Solution - Ex. 7.1.1

a) C = 16.75t + 36,500

b) R = 27t

c) P = 10.25t – 36,500

d) t ≈ 3561 hours (graph and approx. where the lines cross)

1.2 – Functions

- Definitions
- Testing for functions
- Evaluating a function
- Domain of a function
- Applications
- Difference Quotients

Definitions

- function –a function is a relationship between two variables such that to each value of the independent variable there corresponds exactly one value of the dependent variable
- domain –the domain of a function is the set of all values of the independent variable for which the function is defined.
- range – the range of a function is the set of all values assumed by the dependent variable

Definitions

- X – independent variable
- Y – dependent variable

Testing for Functions

- If you are given the points check to see if there are any of the same x-values.
- If so, then it is not a function.
- Easiest way to test if a line is a function is to graph it and do the vertical line test.
- Solve for y
- Graph the line
- Do the vertical line test (if only touches once then it is a function)

Evaluating a function

- This is where we are putting something in our bucket (the variable).
- At a given x-value what is the y-value?

Example 1.1.2

Pg. 19 Ex. 3 – Evaluating a Function

Let g(x) = -x2 + 4x + 1.

Find: (a) g(2)

(b) g(t)

(c) g(x+2).

Solution - Ex. 1.1.2

a. Replace x with 2 in g(x):

g(2) = -(2)2 + 4(2) + 1

= 5

b. Replace x with t in g(x):

g(t) = -(t)2 + 4(t) + 1

= -t2 + 4t + 1

Solution - Ex. 1.1.2

c. Replace x with (x + 2) in g(x):

g(x + 2) = -(x + 2)2 + 4(x + 2) + 1

= -(x2 + 4x + 4) + 4x + 8 + 1

= -x2 + 5

Piecewise-Defined Function

- A function that is defined by two or more equations over a specified domain.
- See example pg. 19 in beige box.

Example 2.1.2

- On the board (19)

Domain of a function

Domain is the set of all real numbers for which the expression is defined.

- It’s as easy as traveling along the x-axis on the road that is your function. You can also figure out where the function cannot be defined.

Applications

- Go over Example 8 on pg. 22

Example 4.1.2

Pg. 29 #89

Find the difference quotient and simplify your answer.

f(x) = x2 – x + 1, (f(2 + h) – f(2))/h, h≠0.

Work on the board!!

Activities (23)

1. Evaluate f(x) = 2 + 3x – x2 for:

a. f(-3)

b. f(x + 1)

c. f(x + h) – f(x)

2. Find the domain: f(x) = 3/(x+1).

1.3 – Graphs of Functions

- The Graph of a Function
- Increasing and Decreasing Functions
- Relative Minimum and Maximum Values
- Graphing Step Functions and Piecewise-Defined Functions
- Even and Odd Functions

The Graph of a Function

- x = the directed distance from the y-axis.
- y = the directed distance from the x-axis.
- Go over Example 2 pg. 31
- Note both Algebraic and Graphical solutions.

Increasing and Decreasing Functions

- Increasing <- the function is rising on the interval
- Decreasing <- the function is falling on the interval
- Constant

Example 1.1.3

Pg. 39 #21

Determine the intervals over which the function is increasing, decreasing, or constant.

f(x) = x3 – 3x2 + 2

Graph on calculator.

Draw on the board.

Relative Minimum and Maximum Values

- A function value f(a) is called a relative minimum of f if there exists an interval

(x1, x2) that contains a such that

x1 < x < x2 implies f(a) ≤ f(x).

- Likewise if x1 < x < x2 implies f(a) ≥ f(x) then f(a) is called the relative maximum.
- See Figure 1.24 on pg. 33

Example 2.1.3

Pg. 39 #31

Use a graphing utility to approximate (to two decimal places) any relative minimum or maximum values of the function.

f(x) = x2 – 6x

Graph on calculator.

Use trace on calculator to approximate.

Even and Odd Functions

- A function is even if, for each x in the domain of f, f(-x) = f(x).
- These graphs are symmetric with respect to the y-axis.
- A function is odd if, for each x in the domain of f, f(-x) = -f(x).
- These graphs are symmetric with respect to the origin.

Example 3.1.3

Pg. 37 Ex. 10

Determine whether each function is even, odd, or neither.

a. g(x) = x3 – x

b. h(x) = x2 + 1

c. f(x) = x3 – 1

Example 4.1.3

Pg. 40 #61

Algebraically determine whether the function is even, odd, or neither.

g(x) = x3 – 5x

Example 5.1.3

Pg. 40 #75

Use a graphing utility to graph the function and determine whether the function is even, odd, or neither.

f(x) = 3x - 2

Solution - Ex. 5.1.3

Pg. 40 #75

The graph is not symmetric to the y-axis or the origin so the function is neither.

1.4 – Shifting, Reflecting, and Stretching Graphs

- Summary of Graphs of Common Functions
- Vertical and Horizontal Shifts
- Reflecting Graphs
- Non-rigid Transformations

Summary of Graphs of Common Functions

- Reading from pg. 42 at the top:
- “One of the goals of this text is to enable you to build your intuition for the basic shapes of the graphs of different types of functions.”
- See the six graphs on pg. 42

Vertical and Horizontal Shifts

- Let c be a positive real number.
- Vertical shift c units upward: h(x) = f(x)+c
- Vertical shift c units downward: h(x) = f(x)-c
- Horizontal shift c units right: h(x) = f(x - c)
- Horizontal shift c units left: h(x) = f(x + c)
- Do the exploration on pg. 43

Example 1.1.4

Pg. 48 #3

Sketch the graphs of the three functions by hand on the same rectangular coordinate system

f(x) = x2

g(x) = x2 + 2

h(x) = (x - 2)2

Reflecting Graphs

- Reflections in the coordinate axes of the graph of y = f(x) are represented as follows.
- Reflection in the x-axis: h(x) = -f(x)
- Reflection in the y-axis: h(x) = f(-x)
- Do the exploration on pg. 45

Non-rigid Transformations

- These are transformations that distort the graph by shrinking and stretching the graph.
- Given by y = cf(x)
- The transformation is a vertical stretch if c>1
- The transformation is a vertical shrink if 0<c<1

Example 3.1.4

Pg. 49 #37

Compare the graph of the function with the graph of f(x) = x3.

p(x) = (1/3x)3 + 2

Example 4.1.4

Pg. 49 #49 and 51

G is related to one of the six parent graph functions on page 42.

- Identify the parent function
- Describe the transformation

1.5 – Combinations of Functions

- Arithmetic Combinations of Functions
- Compositions of Functions

Arithmetic Combinations of Functions

- Let f and g be two functions with overlapping domains. Then, for all x common to both domains, the sum, difference, product, and quotient of f and g are defined as follows:
- Sum: (f+g)(x) = f(x) + g(x)
- Difference: (f-g)(x) = f(x) - g(x)
- Product: (fg)(x) = f(x) · g(x)
- Quotient: (f/g)(x) = f(x) / g(x), g(x) ≠ 0.

Compositions of Functions

- The composition of the funciton f with the function g is:

(f o g)(x) = f(g(x)).

Example 2.1.5

Pg. 59 #39

Find (f o g), (g o f), and the domain of (f o g).

f(x) = sqrt(x + 4), g(x) = x2

Solution - Ex. 2.1.5

Pg. 59 #39

(f o g)(x) = sqrt(x2 + 4)

(g o f)(x) = x + 4, x ≥ -4

Domain = all real numbers

Example 3.1.5

Pg. 57 Ex. 11

- Is the N(T(t)) composition
- Here we just substitute 2 into our composite function in part a
- Solving for t

1.6 – Inverse Functions

- Inverse Functions
- The Graph of an Inverse Function
- The Existence of an Inverse Function
- Finding Inverse Functions Algebraically

Inverse Functions

- When a function f is composed with f-1 (called f-inverse) and vice versa they are equal to x.

The Graph of an Inverse Function

- Reading from pg. 65 (top of the page):
- “The graphs of a function f and its inverse function are related to each other in the following way. If the point (a,b) lies on the graph of f, then the point (b,a) must lie on the graph of f-inverse.”
- The two will be reflections of each other across y=x.
- See Figure 1.68 pg. 65

The Existence of an Inverse Function

- A function f is one-to-one if, for a and b in its domain, f(a) = f(b) implies that a = b.
- A function f has an inverse function if and only if f is one-to-one.

An easy test for one-to-one

- A function is one-to-one if it passes the horizontal line test.
- Two types of functions pass this test
- If f is increasing on its entire domain, then f is one-to-one.
- If f is decreasing on its entire domain, then f is one-to-one.

Example 2.1.6

Pg. 70 #39

Use a graphing utility to graph the function and use the horizontal line test to determine whether the function is one-to-one and so has an inverse.

h(x) = sqrt(16 – x2)

Solution - Ex. 2.1.6

Pg. 70 #39

Graph looks like a rainbow and does not pass the horizontal line test, therefore it is not one-to-one.

Finding Inverse Functions Algebraically

- Use horizontal line test to decide whether f has an inverse function.
- Interchange x and y, and solve for y.
- Verify that the domain of f is equal to the range of f-inverse and f(f-1(x)) = x.

Example 3.1.6

Pg. 70 #59

Find the inverse function of f. Use a graphing utility to graph both f and f-inverse in the same viewing window.

f(x) = 2x - 3

Example 4.1.6

Pg. 70 #65

Find the inverse function of f. Use a graphing utility to graph both f and f-inverse in the same viewing window.

f(x) = sqrt(4 – x2), 0 ≤ x ≤ 2

1.7 – Linear Models and Scatter Plots

- Scatter Plots and Correlation
- Fitting a Line to Data

Scatter Plots and Correlation

- When we graph a set of ordered pairs from a data set we call the collection of points a scatter plot.
- We use these to detect relationships (linear, quadratic, etc.)
- Correlation is a way to describe a positive or negative relationship between the variables.
- See Figure 1.77 on pg. 74

Example 2.1.7

Pg. 78 # 3-6

Determine whether there is positive correlation, negative correlation, or no discernable correlation between the variables.

Solution - Ex. 2.1.7

Pg. 78 #3 - 6

3. Negative correlation

4. No correlation

5. No correlation

6. Positive correlation

Fitting a Line to Data

- Those of you who took or will take a statistics class will cover this in detail. We will do example 5 on page 77.
- I want you to know how to use the regression feature of the calculator.

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