North Carolina State University

North Carolina State University MA 141 Course Text Calculus I by Brenda Burns-Williams and Elizabeth Dempster August 7, 2014

Section1 Functions Introduction In this section, we will define the mathematical meaning of the term function. A function is a special type of relationship between two sets. A function can be represented in a variety of ways; verbally, graphically, numerically, and algebraically. In this beginning module we will study the verbal, numerical and algebraic ways they can be stated. Functions may be thought of as a list of ordered pairs. When these ordered pairs are plotted on a Cartesian Coordinate plane, we get pictures of the functions called graphs. Each of the points on the graph of a function f is of the form (x, f(x)). It is important to keep this in mind as we look at different examples. Knowing which coordinate represents the independent (x) and dependent (f(x)) variables is extremely helpful. In this module the student will learn to work with functions that are defined differently over various parts of their domain. A function that is defined by two (or more) equations over a specified domain is called a piecewise function. After completing this section the learner will be able to do each of the following. Objective 1.1 Use the definition of a function to identify when a relationship is a function. Objective 1.2 Evaluate a function. Objective 1.3 Simplify the difference quotient for a function. Objective 1.4 Find the domain of a function. Objective 1.5 Form the sum, difference, product and quotient of functions. Objective 1.6 Find the domain of the sum, difference, product and quotient functions. Objective 1.7 Identify the graph of a function. Objective 1.8 Identify the domain and range of a function given its graph. Calculus I 1 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.9 Determine f(x) when the graph of f is given. Objective 1.10 Determine where a function is increasing, decreasing, and/or constant by looking at its graph. Objective 1.11 Identify the local maximum and minimum values of a function by looking at its graph. Objective 1.12 Identify and sketch the basic graphs of the following functions: linear, quadratic, cubic, square root, cube root, and absolute value Objective 1.13 Graph a piecewise defined function. Objective 1.14 Determine the value of a piecewise defined function. Objective 1.15 Find a piecewise defined function, f, that fits a given graph. Objective 1.16 Use piecewise defined functions in word problems. Objective 1.17 Graph functions using shifts, reflections, stretches and compressions. Objective 1.18 Find a point on the graph of a transformed function given a point on the original function. Objective 1.19 Determine the function obtained from a series of transformations. Objective 1.20 Define the operation of composition of functions. Objective 1.21 Evaluate a composite function given a table of values. Objective 1.22 Evaluate a composite function given a formula. Objective 1.23 Find a composite function and its domain. Objective 1.24 Find the components of a composite function. Objective 1.25 Determine whether a function is one-to-one by using the horizontal line test. Objective 1.26 Find the inverse of a function when given a table of values or a list of ordered pairs. Objective 1.27 Sketch the graph of f 1 given the graph of f. Objective 1.28 Find f 1 (x), given f(x). Objective 1.29 Find f 1 when f is a domain-restricted function. Calculus I 2 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.1 The Definition of a Function A function is a pairing between two sets of objects for which each element from the first set is paired with exactly one element from the second set. To illustrate, let s take a look at several familiar examples of ways we pair elements from different sets. Let the first set (the domain) be the collection of students: John, Mary, Kim, and Jessica. Let the second set (the range) be the collection of their phone numbers: 555-1234, 555-1235, 555-1245, 555-4321, and 555-3456. When we define the correspondence between the sets to be match the student with his/her phone number(s), we get the following pairings: John 555-1234 Mary 555-1235 555-1245 Kim 555-4321 Jessica 555-3456 Notice that Mary is paired with two phone numbers. Therefore, this pairing is not a function. However, if we let the first set be the same collection of students and the second set be the number of credit hours they are taking this semester, we get the following pairings: John 12 Mary 15 Kim 17 Jessica Each of the students is paired with only one number of credit hours, so this pairing is a function. Notice that both Mary and Jessica are taking 15 credit hours. It s okay for different elements in the first set to be paired with the same element in the second set. The key is to make sure that each element in the first set is paired with only one element in the second set, even if it is the same one each time. There are several special types of functions that are given specific names. Two such functions are: a.) constant function - a function such that every element x X is paired with the same element a Y, i.e. f(x) = a for every element x in the set X b.) identity function - the function such that every element x X is mapped to itself, i.e. f(x) = x Calculus I 3 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.1.1 One way functions can be represented is by ordered pairs, (a, b), which show that the function maps the element a in set X to the element b in set Y. We call a and b coordinates. Determine whether the following sets of ordered pairs are functions. a.) {(1, 4), (2, 3), (1, 7)} This is not a function because 1 is paired with both 4 and 7. b.) {(1, 2), (0, 5), ( 1 2, 3)} This is a function. Each first coordinate has a unique second coordinate. Calculus I 4 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.1.2 Functions can also be represented using data tables, tables that list elements in X and their associated elements in Y. Determine whether each of the following data tables represents a function. a.) x -2 0 0 5 y 5 3 4-1 This is not a function: x = 0 is paired with both y = 3 and y = 4. b.) x -2-1 0 1 2 y -4-2 0 2 4 This is a function. c.) x -2 0 2 4 f(x) 3 3 3 3 This is a function, as each x-value is paired with only one f(x) value. It is an example of a constant function, because each of the elements from the first set is paired with the same element in the second set. Calculus I 5 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.1.3 Decide whether or not y is a function of x: a.) x 2 + y 2 = 9 Recall that y is a function of x only if each x-value corresponds to a unique y-value: y 2 = x 2 + 9 y = ± x 2 + 9 Test to see if y is a function of x now by letting x = 1. Then y = ± 8, but this means that the points (1, 8) and (1, 8) are both valid. This contradicts the definition of a function. So in this case, x 2 + y 2 = 9 is not a function. b.) y = 4x 2 + 3x + 1. This is a function of x. 2x c.) y = 3x 2 27. This is a function of x. Calculus I 6 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.1.4 The function f(x) = x is called the function: a.) constant b.) real c.) identity The function f(x) = x maps each element to itself; thus, f is the identity function. Calculus I 7 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.2 Evaluating Functions Suppose we are given the following data: x 2 1 0 1 2 y 4 2 0 2 4 We see that each of the x-values is paired with a y-value that is two times the value of x. This means that we can find a formula for finding the y-values, namely y = 2x, for each of the x-values listed. It is not always possible to find a function to describe every pairing that we come across; however, this class will deal mainly with functions that have explicit formulae. Both values in the pairings are equally important. To keep up with the pairings between the x and y values we find when we use the formula, a new notation must be used. This notation is called function notation. Instead of naming the second element in the pairing y, we will call it f(x) where f is the function that describes the collection of pairing. This way, when we see f(2) = 4, we know that the value 2 is being mapped to the value 4 and that the ordered pair (2, 4) is an element of f. In the example above, we can define the function given in the table by using the following: f(x) = 2x, where x is a member of the set { 2, 1, 0, 1, 2} In other words, we are now listing the elements of the first set, then giving instructions on how to find the element from the second set with which it is paired. We can then recreate the table of values, list of ordered pairs, or even make a graph of the function. We simply rename y in the chart above as f(x). When we are asked to evaluate f( 1), we can either look at the chart and find the corresponding element for x = 1 or we can use the formula and replace each of the x s in the formula with 1 and then simplify: f(x) = 2x f( 1) = 2( 1) = 2 Using a formula is useful when a function has a large number of data points. Calculus I 8 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.2.1 Let f(x) = 2x 2 + x 1. Evaluate the following: a.) f(0) f(0) = 2(0) 2 + (0) 1 = 2(0) + 0 1 = 0 + 0 1 = 1 b.) f( x) f( x) = 2( x) 2 + ( x) 1 = 2x 2 x 1 c.) f(x) f(x) = ( 2x 2 + x 1) = 2x 2 x + 1 d.) f(a) f(a) = 2(a 2 ) + (a) 1 = 2a 2 + a 1 e.) f(x + h) f(x + h) = 2(x + h) 2 + (x + h) 1 = 2(x 2 + 2xh + h 2 ) + (x + h) 1 = 2x 2 4xh 2h 2 + x + h 1 Calculus I 9 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.2.2 Let h(w) = w 4 w 2 + 1. Evaluate the following: a.) h(2) h(2) = (2) 4 (2) 2 + 1 = 16 4 + 1 = 13 b.) h( 1) h( 1) = ( 1) 4 ( 1) 2 + 1 = 1 1 + 1 = 1 c.) h( w) h( w) = ( w) 4 ( w) 2 + 1 = w 4 w 2 + 1 d.) h(2a) h(2a) = (2a) 4 (2a) 2 + 1 = 16a 4 4a 2 + 1 Calculus I 10 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.2.3 What function corresponds to each of the following data tables? a.) x -2-1 0 1 y 5 5 5 5 Notice that this is a constant function; the function evaluated at any x-value produces the same value every time. Hence, our function is f(x) = 5. b.) x -2-1 0 1 2 y -14-7 0 7 14 All of the y-values are multiples of 7. Notice that if we multiply each x-value by 7, we obtain its corresponding y-value. Hence, our function is f(x) = 7x. Calculus I 11 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.3 The Difference Quotient for Functions The difference quotient for a given function f(x) is the expression f(x + h) f(x), h 0. h In calculus you will see that this expression is used as a way of calculating the slopes of special lines called secant lines. Finding the slopes of secant lines plays a valuable role is determining how the output of a function is changing as well as how quickly it is changing. For now, we will concentrate on simplifying this expression for given functions so that this part of the process will be easy when you take calculus. Let f(x) = x 2 3x. To simplify the difference quotient, we will need to evaluate f(x+h). Caution: To evaluate f(x + h) you must replace all the x s in f(x) with x + h. The CORRECT way is f(x + h) = (x + h) 2 3(x + h) The INCORRECT way is which is just adding h to f(x). f(x + h) = x 2 3x + h, So, the difference quotient for f(x) = x 2 3x is: f(x + h) f(x) h = [(x + h)2 3(x + h)] [x 2 3x] h = x2 + 2xh + h 2 3x 3h x 2 + 3x h = 2xh + h2 3h h h(2x + h 3) = h = 2x 3 + h Notice that we had to distribute negatives - it is easy to miss this step! Calculus I 12 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.3.1 Find and simplify the difference quotient for the function f(x) = 3x 2 + 5x. f(x + h) f(x) h = [3(x + h)2 + 5(x + h)] [3x 2 + 5x] h = [3(x2 + 2xh + h 2 ) + 5x + 5h] [3x 2 + 5x] h = 3x2 + 6xh + 3h 2 + 5x + 5h 3x 2 5x h = 6xh + 3h2 + 5h h = h(6x + 3h + 5) h = 6x + 3h + 5 Calculus I 13 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.3.2 Let g(x) = w + 2. Find and simplify the difference quotient g(w + h) g(w) h g(w + h) g(w) (w + h) + 2 w + 2 =, h h This is the most simplified form, because you cannot combine the square roots or cancel out the h terms. Calculus I 14 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.4 The Domain of a Function The domain of a function is the list of values that are used from the first set of elements X to make the pairing. When given the function in ordered pair or table of values form, it is easy to find the domain by looking at the list of x-values given. However, when given the formula for the function and asked to find a domain, you are actually being asked to find the largest set of real numbers such that if you evaluate the function at those numbers you will get a real number. In other words, the domain of a funtion is the largest set of real numbers for which the function is defined. Let s look at some examples: f(x) = x 3 + 2x, g(x) = 5x x 2 9, h(x) = t 2 25 The domain of f is all real numbers (R), also denoted as (, ), since any input generates a real number. To find the domain of g, we need to check where the denominator equals zero, since fractions with zeros in the denominator are undefined. Thus, we solve the equation x 2 9 = 0 to find the numbers that do not work: x 2 9 = 0 x = 3, 3 Thus, the domain of g is (, 3) ( 3, 3) (3, ). This says that we can pick any real number except 3 and 3 to get a real number from the function g. In our interval notation, if we use a parenthesis, ( or ), we do not include the endpoint numbers in our interval, but if we use a bracket, [ or ], we do include those endpoints. For our third example, h(x), we know that it is not possible to take the square root of a negative number and get a real number. Hence, the domain is all real numbers such that the inside piece is greater than or equal to zero (since 0 = 0 is defined). Thus, t 2 25 0 (t + 5)(t 5) 0 This is only true in two cases: a.) both components are nonnegative, i.e. b.) both components are nonpositive, i.e. t + 5 0 and t 5 0 t 5 and t 5 t + 5 0 and t 5 0 t 5 and t 5 In the first case, the pair of inequalities simplifies to t 5; in the second case, t 5. Thus, the domain of h is {x x 5, x 5}, or in interval notation, (, 5) (5, ). Calculus I 15 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.4.1 Find the domain of each of the following: a.) g(x) = x + 4 x 3 4x g(x) is undefined when the denominator x 3 4x = 0: x 3 4x = 0 x(x 2 4) = 0 x(x + 2)(x 2) = 0 So, x = 0, x + 2 = 0, and x 2 = 0. Therefore, the x-values that give a zero in the denominator are x = 0, x = 2, x = 2. The domain of g(x) is all real numbers except 0, 2, 2, i.e. {x R x 0, 2, 2}, or in interval notation, (, 2) ( 2, 0) (0, 2) (2, ). 4 b.) h(x) = x 9 We need to check two cases here: (a) when the denominator is zero, i.e. x 9 = 0 x 9 = 0 (b) when x 9 is undefined, i.e. x 9 0. If we put these two cases together, the good x-values are when x 9 > 0 x > 9 Hence, the domain of h(x) is {x R x > 9} or in interval notation, (9, ). c.) j(x) = 5x 3 + 4x 2 + 1 Notice that you can evaluate j(x) at any real number and the function will generate a real number. Hence, the domain for j(x) is all real numbers, i.e. R or (, ). d.) e(x) = 3 x Notice that e(x) is an odd root; so the domain is all real numbers, R or (, ). We only have problems with even roots; specifically, we cannot take the even root of a negative number. To check this issue with odd roots, just evaluate e(x) at 0 and at any negative number: 3 8 = 2, 0 = 0. Calculus I 16 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.4.2 Find the domain of f(x) = 1 x 5 + x. Remember, first look for the problem spots. Notice that since f(x) has both a denominator and a square root, we will have restrictions on what values can be in the domain. The first term of the function, 1 x 5, tells us that x 5 0 x 5, because the denominator cannot be 0. The second term of the function, x, implies that x 0, because the inside piece of the square root cannot be negative. If we put these two pieces together, x 5 and x 0, we can describe the domain as follows: or in interval notation, {x R x 0, x 5} [0, 5) (5, ). Objective 1.5 Find the sum, difference, product and quotient of functions. Given that f and g are functions, then the general definitions are: a.) (f + g)(x) := f(x) + g(x) b.) (f g)(x) := f(x) g(x) c.) (f g)(x) := f(x) g(x) d.) ( f g ) (x) := f(x) g(x), g(x) 0 Calculus I 17 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.5.1 Let f(x) = x + 5 and g(x) = x 2 1. Find each of the following: a.) (f g)(x) (f g)(x) = f(x) g(x) = (x + 5) (x 2 1) = x + 5 x 2 + 1 = x 2 + x + 6 b.) ( ) f (x) g ( ) f (x) = f(x) g g(x) = x + 5 x 2 1 c.) ( ) f (2) g ( ) f (2) = g 2 + 5 2 2 1 = 7 4 1 = 7 3 Calculus I 18 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.5.2 The tables for f and g are as follows: x -1 0 1 2 f(x) 5-8 0 1 x -1 0 1 3 g(x) 3 6-2 5 Find the following: a.) (fg)(0) (fg)(0) = f(0) g(0) = ( 8)(6) = 48 b.) (f + g)(2) Looking at the two tables, we see that 2 is not in the domain of g. Therefore, (f + g)(2) = f(2) + g(2) = 1 + undefined = undefined. c.) ( ) g (1) f ( ) g (1) = g(1) f f(1) = 2 0 = undefined. Calculus I 19 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.6 Finding the domains of the sum, difference, product, and quotient functions The domain of the sum, difference, and product of two functions is equal to the set of numbers x that are in both the domain of f and of g. In other words, if we let D(f) represent the domain of the function f, then D(f + g) = D(f g) = D(fg) = D(f) D(g). For the quotient of two functions, we also have the restriction that g(x) 0. Hence, the domain for the quotient of two functions is the set of numbers x that are in both the domain of f and of g minus the set of numbers x where g(x) = 0. In other words, or D( f ) = [D(f) D(g)] {x R g(x) = 0} g D(f) {x D(g) g(x) 0}. Calculus I 20 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.6.1 If the domain of f is all real numbers in the interval [0, 7] and the domain of g is all real numbers in the interval [ 2, 5], find the domain of f + g. We can use a number line to illustrate the two domains. We are looking for the set of numbers x such that x is in both the domain of f and the domain of g. Using our figure, we can see that the number lines overlap on the interval [0, 5], so the domain of f + g is [0, 5]. Calculus I 21 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.6.2 Let f(x) = x x + 3 and g(x) = x x 2. Find ( ) f (x) and its domain. g First, we need to find the domain of f and the domain of g. Generally, if the domain is not specified, then we begin with the set of all real numbers R and then remove the numbers that cause problems (for example, the numbers that make the denominator equal to zero or that result in taking the square root of negative numbers). The domain of f is D(f) = {x (R) x 3}, or (, 3) ( 3, ), and the domain of g is D(g) = {x R x 2}, or (, 2) (2, ). Therefore, we know that the domain of the quotient of the two functions is at most {x R x 3, 2}. Since we are working with the quotient of two functions, we also need to find the x-values for which g(x) = 0. So, set the function g equal to zero and multiply each side of the equation by the denominator, x 2: g(x) = x x 2 = 0 x = 0. Note that when working with fractions, we set the denominator equal to zero to find values for which the function is undefined and we set the numerator equal to zero to find values for which the function is zero. We can use number lines to visualize this information: The red open circles represent values not in the domain of the functions. Recall that Hence, D(f/g) = [D(f) D(g)] {x R g(x) = 0}. D(f/g) = {x R x 3, 2} {0} = {x R x 3, 0, 2}. Calculus I 22 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

( ) Now, let s find f g (x): ( ) f (x) = g = x x+3 x x 2 ( x x + 3 = x 2 x + 3. ) ( ) x 2 x Calculus I 23 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.6.3 Given h(x) = 2x 1 and j(x) = x 2 + x 2, a.) Find the difference function (h j)(x) and its domain. (h j)(x) = (h(x) j(x)) = (2x 1) (x 2 + x 2) = 2x 1 x 2 x + 2 = x 2 + x + 1 D(h) = R and D(j) = R, so D(h j) = R R = R, or (, ). b.) Find ( ) h (x) and its domain. j ( ) h (x) = j = 2x 1 x 2 + x 2 (2x 1) (x 1)(x + 2) We can find the domain of the quotient function using the following formula: ( ) h D = D(h) D(j) {x R j(x) = 0}. j We know that: (a) D(h) = R (b) D(j) = R (c) j(x) = 0 (x + 2)(x 1) = 0 x = 2 or x = 1 Hence, D ( ) h j = R R { 2, 1} = R { 2, 1} = {x R x 2, 1} = (, 2) ( 2, 1) (1, ). Calculus I 24 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.7 Identify the graph of a function We can make a picture of a function by plotting its ordered pairs (x, f(x)) on a Cartesian Coordinate plane. When we graph a function, no two points on the graph should be on the same vertical line. Otherwise, the function would contain two points with the same x-coordinate but two different y-coordinates. This idea is the basis for the Vertical Line Test, a well-known test to determine whether or not a graph is a function. To use this test, look at the graph and determine whether any vertical line you might draw would intersect the graph at more than one point. If this can be done, then the graph fails the vertical line test and the graph is not a function. If no such vertical line can be drawn then the graph passes the vertical line test and it is a function. Figure 1.7.1: A Function Figure 1.7.2: Not a Function The graph in Figure 1 is a function, because it passes the vertical line test; no matter where we move the vertical line, it will only ever cross the graph at one point. However, the graph in Figure 2 fails the function, because the vertical line crosses the graph at two points; hence, this graph is NOT a function. Calculus I 25 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.7.1 Is the following the graph of a function? This graph fails the vertical line test, therefore the graph is not a function. Calculus I 26 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.8 Identify the domain and range of a function when given its graph. For example, consider the graph of f(x) = x 2. The domain is the set of x-values. From the picture we see that every real number is used to get the graph of f. Therefore, the domain of f is all real numbers, i.e. R or (, ). By looking at the graph of f, we also note that the only y-values that are used to make the graph are values greater than or equal to 0. Therefore, the range of f is y 0. To express the range in interval notation we write [0, ). Calculus I 27 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.8.1 Given the graph of g below, determine the domain and the range. The domain of g is [ 4, 5] and the range of g is [ 3, 3]. Calculus I 28 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.8.2 Given the graph of f below, find the domain and range. The domain of f is [ 5, ) and the range is [ 3, ). Calculus I 29 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.9 Determine the best graphical representation for a given word problem. In mathematics, we often want to draw a picture to represent data that has been described in words. Before we discuss a method that will help us represent written data, we need to define a few terms: a.) independent variable- the input value type b.) dependent variable- the output value type; depends on the input For example, for the function f(x) = x 2, the independent variable is x and the dependent variable is y = f(x). Method: a.) Consider what is being modeled. b.) Determine what the independent variable should represent: time, weight, height, etc. c.) Determine what the dependent variable should represent: speed, cost, etc. d.) When examining a choice of graphs, consider how the independent and dependent variables are related. For example, if you were asked to describe the height of a person as a function of time, you would not expect the graph to fluctuate up and down, nor to decrease. Rather you would expect that the person would grow at a fairly steady rate until he reached maturity at which point you could expect his height to remain fairly constant. So, the graph would look something like: The independent variable in the above example would be time and height would be the dependent variable (height depends on how much time has passed). Calculus I 30 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.9.1 Match each description with the graph that best represents it. a.) The height of the tide as a function of time. b.) The cost of mailing a letter as a function of its weight. c.) The speed of a car traveling through town as a function of time. (b) (c) (d) 1 a, 2 b, 3 c Calculus I 31 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.10 Determine where a function is increasing, decreasing, and/ or constant by looking at its graph. We say that a function is increasing on an open interval I = (a, b) if for any choice of x 1, x 2 in I, f(x 1 ) < f(x 2 ). Visually, the graph of the function appears to be rising as you travel from left to right along the graph. Similarly, a function is decreasing on an open interval I = (a, b) if for any choice of x 1, x 2 in I, f(x 1 ) > f(x 2 ); here, the graph appears to be falling as you traverse from left to right along the function. When the graph of a function is horizontal, namely that f(x 1 ) = f(x 2 ), then the function is said to be constant. Calculus I 32 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.10.1 Let g be given by the graph below. Determine where g is increasing, decreasing, or constant. The graph is rising from the point ( 4, 0) to the point ( 2, 3), and also from the point (2, 3) to the point (4, 0); so we say the function g is increasing on the interval ( 4, 2) (2, 4). The graph is falling from the point ( 2, 3) to the point (2, 3). Therefore, we say that the function g is decreasing on the interval ( 2, 2). The graph is horizontal from the point (4, 0) to the point (5, 0). Hence, the function g is constant on the interval (4, 5). Calculus I 33 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.11 Identify the local maximum and minimum values of a function by looking at its graph. A function f(x) has a local maximum at a point x = c if and only if there exists some open neighborhood I containing c such that f(c) f(x) for all x in I. Here, an open neighborhood is just an open interval (a, b) containing c. This means that there must be points on either side of the value x = c. Hence, endpoints on a graph cannot be local maximums. We say that f(c) is the local maximum and c is where it occurs. Similarly, a function f(x) has a local minimum at a point c if and only if there exists some open neighborhood I containing c such that f(c) f(x) for all x in I. Again, I is just an open interval containing c, and hence, endpoints on a graph cannot be local minimums either. By definition, constant portions of the graph, with the exception of the endpoints, are both local maximums and local minimums. For example, let s use a quadratic function, f(x) = x 2. This function has a local minimum of 0 at x = 0, since 0 is the smallest value y attains in the range of f. Notice that if we restricted the domain of f to be {x R x 0}, we get the following graph: This function DOES NOT have a local minimum, because (0, 0) is an endpoint. Calculus I 34 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Now let g(x) = x 2 : This function has a local maximum of 0 at x = 0, since 0 is the largest value y attains in the range of g. Again, if we restricted the domain of g to {x R x 0} or any other set that either makes (0, 0) an endpoint or excludes (0, 0), then wouldn t have a local maximum. Calculus I 35 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.11.1 Find the local maximum(s) and minimum(s) of the following function: There are two local maximums. The first maximum is f(x) = 3, occurring at the point ( 2, 3). The second local maximum is at f(x) = 0, occurring on the interval (4, 5), the constant portion of the graph. There are two local minimums. The first local minimum is f(x) = 3, occurring at the point (2, 3). The second local minimum is at f(x) = 0, occurring on the interval (4, 5), the constant portion of the graph. Calculus I 36 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.11.2 Find the local maximum and the local minimum, and state where the function is increasing or decreasing. The local maximum of the function is 0. The local minimum of the function is 4. Note that this is not the smallest value that the function takes on, but is the smallest in an open neighborhood of the point (0, 4). The function is increasing on the interval (, 2) (0, ). The function is decreasing on the interval ( 2, 0). Calculus I 37 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.12 Identify and sketch the basic graphs of the following functions: linear, quadratic, cubic, square root, cube root, absolute value, and reciprocal. Each of the following graphs is important for you to have in your memory banks. You will need to memorize how each of them look. You will also need to be able to state the domain, range, intervals where the function is increasing, decreasing, or constant as well as find any local maximums or minimum values. In other words, you need to know everything about them! a.) The Basic Linear Graph: y = x Domain: (, ) Range: (, ) Increasing: (, ) Decreasing: None Local Max: None Local Min: None Calculus I 38 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

b.) The Basic Quadratic Graph, or Parabola: y = x 2 Domain: (, ) Range: [0, ) Increasing: (0, ) Decreasing: (, 0) Local Max: None Local Min: y = 0 at x = 0 Calculus I 39 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

c.) The Basic Cubic Graph: y = x 3 Domain: (, ) Range: (, ) Increasing: (, ) Decreasing: None Local Max: None Local Min: None Calculus I 40 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

d.) The Basic Square Root Graph: y = x Domain: [0, ) Range: [0, ) Increasing: (0, ) Decreasing: None Local Max: None Local Min: None Calculus I 41 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

e.) The Basic Cube Root Graph: y = 3 x Domain: (, ) Range: (, ) Increasing: (, ) Decreasing: None Local Max: None Local Min: None Calculus I 42 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

f.) The Basic Absolute Value Graph: y = x Domain: (, ) Range: [0, ) Increasing: (0, ) Decreasing: (, 0) Local Max: None Local Min: y = 0 at x = 0 Calculus I 43 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

g.) The Basic Reciprocal Graph: y = 1 x Domain: (, 0) (0, ) Range: (, 0) (0, ) Increasing: None Decreasing: (, 0) and (0, ) Local Max: None Local Min: None Calculus I 44 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.13 Graph a piecewise defined function To sketch the graph of a piecewise function, we can first graph each component of the function and then combine the graphs based on the domain of each portion. For example, suppose that x + 1, 1 x < 1 f(x) = 2, x = 1. x 2, x > 1 Each of these graphs can be drawn separately, as seen below. The only thing that we have to do is take pieces of each graph and combine them together. Piecewise graphs are not necessarily continuous, so there might be values that jump from one y-value to another at a given x-value. (a) y = x + 1 (b) y = 2 (c) y = x 2 Considering the restrictions given in the function f(x), we can put all three graphs together to get the final piecewise graph: Calculus I 45 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.13.1 Sketch the graph of the following function. 3 + x, 3 x < 0 g(x) = 3, x = 0 x, x > 0 We can do this in parts by sketching graphs of each of the three smaller functions in the definition of the piecewise function and putting them together to find the graph of g(x). (b) (c) (d) Again, considering the domain restrictions specified in the function g(x), we can put the above graphs together to yield the graph of g(x): Calculus I 46 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.13.2 Sketch the graph of the following piecewise function: 5x, x < 4 f(x) = 25, x = 4 x 2 + 4, x > 4. We can do this in parts by sketching graphs of each of the three smaller functions in the definition of the piecewise function and putting them together to find the graph of f(x). (b) (c) (d) Putting these together yields the following graph of f(x): Calculus I 47 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.14 Determine the value of a piecewise defined function To determine the value of a piecewise function f at a given point x, we first need to determine which component of the piecewise function has a domain that includes x. If x is not in the domain of any of the components, then f(x) does not exist. However, if x is in the domain of a component, then we use this equation to find the value of f(x) by plugging in x. Consider the function 5x, x < 4 f(x) = 25, x = 4. x 2 + 4, x > 4 Suppose we want to find the value of f(x) at x = 4. We first need to find the component of f(x) that is defined when x = 4, i.e. the function y = 25. We then use the equation y = 25 to find the value of f(x) at x = 4; therefore, the value of f(4) = 25. If we let x = 5, we would use the function y = x 2 + 4 instead. Hence, f(5) = 5 2 + 4 = 29. We can confirm these values by looking at the graph of f(x) depicted below: Calculus I 48 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.14.1 Suppose that 3 + x, 3 x < 0 g(x) = 3, x = 0 x, x > 0 Find g( 1). When x = 1, we need to look at the function y = 3 + x. So g( 1) = 3 + ( 1) = 2. Looking at the graph confirms this. Calculus I 49 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.15 Find a piecewise function for a given graph. Sometimes, when given a graph of a function, it will be best described by a piecewise function. There are several steps to follow that make finding the function easier. a.) Determine the number of pieces into which the graph can be broken. b.) Find the domain of each of the pieces, being careful that the domains do not overlap, but that all values in the domain of the graph are included. Pay special attention to the points where the graph is discontinuous or where values do not exist. c.) Determine the form of each of the pieces. What are the parent functions? d.) For each portion, try to come up with a modification of a parent function that will describe the line. The graphs that you will see in this section will have easy modifications of the parent functions. e.) Put all of these pieces with their domains and create the function that describes the entire graph. To check the end result, graph each component function individually over its normal domain and combine the graphs, using your function to define the domain for each component. If the combined graph matches the image given, then your function is correct. Calculus I 50 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.15.1 Given the following sketch of the function h(x), find the explicit formula for the function. In the above graph, we see a linear component on the interval (, 1] and a constant component on the interval (1, ). Because we have two points, (1, 4) and (0, 3), and we know that the y-intercept is at the point (0, 3), we can use the slope-intercept formula to find the equation of the linear component of our graph. The slope-intercept formula is y = mx + b, where m is the slope and b is the y-intercept. We just need to find the slope m: m = y 2 y 1 x 2 x 1 = 3 4 0 1 = 1 1 = 1 Therefore, our equation for the linear piece on the interval (, 1] is y = x + 3. On the interval (1, ) we can see that the line is constant (and therefore has a slope of 0) at the value y = 1; thus, the equation for this line is y = 1. Putting all of this information together, we see that the piecewise function h(x) is defined to be x + 3, x 1 h(x) =. 1, x > 1 Calculus I 51 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.16 Use piecewise functions in word problems. This objective discusses creating a piecewise function from a word problem: In the United States, we have a graduated income tax, i.e. the amount of tax that you pay is a percentage of your total income. Suppose that during a particular year the following rates hold: the first tier is on income less than $17,900 with a payment of 15% tax, the second tier requires that you pay the tax from the previous tier plus an additional 28% tax on income between $17,900 and $43,250, and the third tier requires that you pay the taxes of the previous tiers plus 31% tax on any income greater than $43,250. The easiest place to break up the word problem is at the income amount that marks the transition between tiers. We also need to note the amount of income that falls in each tier by subtracting the amount at the bottom of the tier from the amount at the top of the tier. Therefore, there are $17,900 in the first tier and $25,350 in the second tier. To pull it all together, we can make a chart of the following information: Piece Tax Start $ End $ Prev tier tax (1) Prev tier tax (2) 1 15% 0 17900 N/A N/A 2 28% 17900 43250 15% on 17900 N/A 3 31% 43250 N/A 28% on 25350 15% on 17900 The start and end dollar amounts make up the domain for each of the pieces of our new function and with careful reading of the word problem, we can get the appropriate inequality signs. The trickiest part of this problem is to make sure to only take tax on the appropriate amounts. The first piece of the function is the easiest: f(x) =.15x, since we do not have to account for any previous tiers. The second piece is more difficult. We first have to take 15% of $17,900 and then 28% on whatever is left; so we have f(x) =.15(17900) +.28(x 17900). Similarly, the third function can be given as f(x) =.15(17900) +.28(43250) +.31(x 43250). Combining this with the domain of each piece, we have the following piecewise function:.15x, 0 x < 17900 f(x) =.28(x 17900) + 2685, 17900 x 43250.31(x 43250) + 9783, x > 43250 Calculus I 52 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.16.1 Write a piecewise-defined function that models the following information about postal rates on first-class mail. If the letter weighs less than or equal to 1 ounce, it costs 37 cents. If the letter weighs more than 1 ounce but less than or equal to 2 ounces, it costs 60 cents to mail. If the letter weighs more than 2 ounces but less than or equal to 3 ounces, it costs 83 cents to mail. Write a function c(x) for the cost depending of the ounces x. Sketch the graph of the function. c(x) = 37, 0 < x 1 60, 1 < x 2 83, 2 < x 3. Notice the restriction placed on x for the very first component of the function c(x); x has a lower bound of 0. When working with word problems, we have to use commonsense. We would never have a letter that weighed negative ounces. So, we must be careful when using units such as ounces, inches, etc. Calculus I 53 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.17 Graph functions using shifts, reflections, stretches and compressions. Horizontal Translations:// In this case, we add, subtract, multiply, divide, or negate numbers all inside the function; these act on the variable x. a.) Shift: y = f(x + c) (a) If c > 0, then the function shifts left c units (b) If c < 0, then the function shifts right c units b.) Stretching/Compression: y = f(cx) (a) If 0 < c < 1, then the graph stretches by a factor of 1 c. (b) If c > 1, then the graph compresses by a factor of 1 c. c.) Reflection about the y-axis: y = f( x). Vertical Translations:// In this case, we add, subtract, multiply, divide, or negate numbers all outside of the function; these act on the entire function f(x). a.) Shift: y = f(x) + c (a) c > 0, shift up c units (b) c < 0, shift down c units b.) Stretching/Compression : y = cf(x) (a) If 0 < c < 1, then the function compresses by a factor of c. (b) If c > 1, then the function stretches by a factor of c. c.) Reflection about the x-axis: y = f(x). Notice that, if done carefully, a horizontal stretch/compression can also be written as a vertical stretch/compression, and vice versa. By rules of functions, f(cx) = cf(x). We must be careful because f(cx + b) cf(x + b). So, in order to change the type of translation, the coefficient of x must be 1; for example, rewrite f(cx + b) as f(c(x + b / c)) = cf(x + b / c). Calculus I 54 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.17.1 For each function given below, state the parent function, draw its graph, state the type of translation involved, and sketch the graph of the translated function. a.) f(x) = (x 2) 3 b.) f(x) = 3 x + 2 c.) f(x) = x d.) f(x) = ( x) 3 The parent function is given in red and the translated function is given in blue. a.) f(x) = (x 2) 3 For f(x) = (x 2) 3, the parent function is y = x 3. Notice that we subtract 2 inside the function, and hence this type of translation is a horizontal shift. Figure 1.17.1: Parent function: f(x) = x 3. Translation: Horizontal shift by 2 units to the right b.) f(x) = 3 x + 2 For f(x) = 3 x + 2, the parent function is y = 3 x. Notice that we add 2 inside the function; thus, this translation is also a horizontal shift. Figure 1.17.2: Parent function: f(x) = 3 x. Translation: Horizontal shift by 2 units to the left Calculus I 55 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

c.) f(x) = x For f(x) = x, the parent function is f(x) = x ; so for this translation, we have negated the entire function. Hence, this is a reflection about the x-axis. Figure 1.17.3: Parent function: f(x) = x. Translation: Reflection about the x-axis. d.) f(x) = ( x) 3 This function also has a parent function of f(x) = x 3, but this time we have negated the variable x, i.e. the inside of the function. Therefore, this translation is a reflection about the y-axis. Figure 1.17.4: Parent function: f(x) = x 3. Translation: Reflection about the y-axis. Calculus I 56 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.17.2 Sketch the graph of y = f(x) + 2 where f(x) = x 2. We know that the parent function f(x) looks like: First, we check for horizontal translations, looking inside the function to see if we have changed our variable in anyway. In this case, everything is outside the function. So, next we check to see if our parent function has been negated and then to see if any real numbers have been multiplied to our parent function. In this case, we have f(x), so we need to reflect the graph x 2 over the x-axis. Finally, we check to see if any real numbers have been added or subtracted. In this case, since we have y = f(x) + 2, we have to shift the graph up two units. Therefore, the final graph looks like: Calculus I 57 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.17.3 Given the graph of f below, sketch the graph of y = 2f( x) 3. Again, first we decide if any translations occur inside the parent function. In this case, we don t have any shifts or stretches/compressions. We only have the function f( x), in which the variable is being negated. Hence, we reflect our graph about the y-axis. Now we can look for translations on the outside of the parent function. Remember, first we look for negation, then multiplication or division, and finally addition and subtraction. For this problem, we have multiplied f( x) by 2, so we have stretched our graph vertically by a factor of 2. Last, since we subtracted 3, we shift our graph down 3 units. (b) Reflect about y-axis (c) Stretch vertically by a factor of 2 (d) Shift down 3 units We will discuss in greater detail how translations affect specific points in the next section, but notice in this example how the points change when we apply certain transformations. When we reflected the initial graph about the y-axis, the x-values for each point were negated. When we stretched the the graph by a factor of 2 in part b), we multiplied the y-values by 2. Finally, when we shifted the graph down by 3 units, we subtracted 3 from the y-values. Calculus I 58 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.18 Find a point on the graph of a transformed function given a point on the original function. Sometimes when given a graph of a function, finding the graph of the translated function can be difficult. One way to get around this difficulty is to look at the translation of individual points. This way you can construct the new graph in steps. Consider the point ( 4, 2) on the graph f(x). Let s assume we don t know the original graph. However, we can still figure out where the point is translated on a new graph. For example, let y = 3f( 2x 4) + 3. We can see that there are six things happening to the graph of f(x): a.) A vertical reflection (negation outside the function), b.) A vertical stretch (multiplication outside the function), c.) A horizontal reflection (negation inside the function), d.) A horizontal compression (multiplication inside the function), e.) A horizontal shift (subtraction inside the function), and f.) A vertical shift (addition outside the function). With so many translations occurring, it is important to remember the order of operations. In order to apply the rules we learned in Objective 5.1, we must first make the coefficient on the x term equal one. Therefore, we have 3f( 2x 4) + 3 = 3f ( 2(x + 2)) + 3, which we obtained by factoring out a 2 from the argument of f(x). We can now see what the values of each of our translations are: a.) A vertical reflection over the x-axis, b.) A vertical stretch by a factor of 3, c.) A horizontal reflection over the y-axis, d.) A horizontal compression by a factor of 1 2, e.) A horizontal shift 2 units to the left, and f.) A vertical shift 3 units up. Remember that horizontal shifts move in the opposite direction of the sign ( + mean move left and - means move right) and that a horizontal stretch or compression is by the reciprocal of the number you see (if a is the value of the horizontal stretch or compression, then the graph stretches or compresses by a factor of 1 a ). By order of operations, we first apply horizontal shifts, and then apply the vertical shifts. Hence, for this example, we first shift horizontally, then compress horizontally, and then reflect over the y-axis. After we have finished manipulating the x-value, we then move onto the y-value. Here, the Calculus I 59 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

order of operations are in reverse of horizontal translations: first we reflect over the x-axis, then we stretch vertically, and finally, we shift vertically. Using these rules, we can see the progression of the point ( 4, 2) under the translation y = 3f ( 2(x + 2)) + 3: ( 4, 2) shift left 2 add -2 to x value ( 6, 2) compress horizontally by a factor of 2 multiply x value by 1 2 ( 3, 2) reflect over vertical axis multiply x value by -1 (3, 2) reflect over horizontal axis multiply y value by -1 (3, 2) stretch vertically by a factor of 3 multiply y value by 3 (3, 6) shift up 3 add 3 to y value (3, 3). Thus, the point ( 4, 2) on the graph of f(x) is translated to the point (3, 3) on the graph y = 3f ( 2(x + 2)) + 3. Calculus I 60 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.18.1 If the point (5, 3) is a point on the graph of f(x), find a point on the graph of y = 2f(x + 1). By applying translations, we can use the point (5, 3), which is on the graph of f(x), to find a point on the graph of y = 2f(x + 1). Since the variable x has a coefficient of 1, we can immediately begin to apply translations to our point. First, notice that the function f(x) has a horizontal shift of 1 unit to the left. Hence, (5, 3) (4, 3). Next, notice that f(x + 1) is stretched vertically by a factor of 2 (changing the y-value); so (5, 3) (4, 3) (4, 6). Therefore, a point on the graph y = 2f(x + 1) is (4, 6), which we obtained by translating the point (5, 3). Calculus I 61 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.18.2 Sketch the graph of f(x) = x + 5. We know the parent function is f(x) = x. First, the function is shifted left by 5 units and then reflected over the x-axis. This yields the following graph: Calculus I 62 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.18.3 Given the graph of f below, sketch the graph of y = 2f( x) 3 by translating the labelled points as described above. refl y-axis vert factor of 2 down 3 ( 3, 2) (3, 2) (3, 4) (3, 1) ( 2, 0) (2, 0) (2, 0) (2, 3) ( 1, 2) (1, 2) (1, 4) (1, 1) (1, 2) ( 1, 2) ( 1, 4) ( 1, 1) Calculus I 63 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.18.4 Consider the sketch of the following function. What is its basic shape? How has that basic shape been translated? Determine a formula for this function. Notice that the basic shape of the function is an upward facing parabola; so the parent function is f(x) = x 2. In the parent function, the vertex of the parabola is at the point (0, 0); here, we can see that the vertex is ( 3, 4). Therefore, we can consider a possible translation to be 3 units to the left and 4 units down, giving us the function y = (x + 3) 2 4. We just need to check at least one more point to make sure that there is not a vertical or horizontal stretch. The point (1, 1) is on the graph f(x) = x 2. Using our translation, this point moves to the point ( 2, 3). If we plug the value 2 in for x into our new formula, we also get the point ( 2, 3). Therefore, our formula is y = (x + 3) 2 4. Calculus I 64 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.18.5 Given the function f(x) = e x, find the function g(x) obtained by performing the following translations: a.) Horizontal shift left 3 units b.) Reflection about the vertical axis c.) Vertical shift up 1 unit The function is translated in the following way: f(x + 3) f( (x + 3)) f( (x + 3)) + 1 Add 3 to the x-value to shift left Reflect the new x-value about the y-axis Shift the output up 1 unit Thus g(x) = f( (x + 3)) + 1 = e (x+3) + 1. The images of the graphs under each translation are below. (b) f(x) (c) f(x + 3) (d) f( (x + 3)) (e) f( (x + 3)) + 1 Calculus I 65 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.19 Define the operation of composition of functions. The composition of two functions is formally defined as (f g)(x) = f( g(x)), where the operator defines composition. This is often read as f composed with g or f of g So what is composition? The idea is to apply one function to your x-value and then apply the second function to the result of the first action, making sure to work from right to left, or from the inside to the outside (depending on your notation). One way to think of this is to let g(x) = y. Then the composition of functions (f g)(x) = f( g(x)) = f(y). The following sections describe how to evaluate and construct composition of functions. Calculus I 66 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.20 Evaluate a composite function given a table of values. Suppose that you are given a table of values for f(x) and g(x). How would you use these values to find the composition (f g)(x)? Remember that the notation (f g)(x) = f( g(x))(x) means evaluate g(x), and then evaluate the answer in f(x). In other words, we want to work from the inside to the outside. Thus, to use a table of values to find (f g)(x), first use the table to evaluate g(x). This value, x becomes your new x-value. Then, evaluate f(x ), which is the solution for the composition of the two functions evaluated at x. Calculus I 67 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.20.1 The table for f is as follows: The table for g is as follows: x 0 1 2 3 f(x) 0 1 4 9 x 1 2 3 4 g(x) 3 6 8 10 Use the above tables to: a.) Find g composed with f at the value 2, i.e. (g f)(2) = g(f(2)). To find (g f)(2), first evaluate the right (or inside) function, f(x), at the value x = 2, i.e. evaluate f(2). Using the table, we know f(2) = 4. Then, evaluate g(x) at this value: g(4) = 10. A summary using equations follows: Therefore, (g f)(2) = 10. (g f)(2) = g(f(2)) = g(4) = 10. b.) Find f composed with g at the value 1, i.e. (f g)(1). Using the tables, we have: (f g)(1) = f(g(1)) = f(3) = 9. Calculus I 68 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.21 Evaluate a composite function given a formula. Evaluating composite functions using formulas is similar to evaluating these functions using tables. The only difference is that algebra is used for the evaluation process. To evaluate a composite function given two formulas and a point, evaluate the inner function at the given point. Then, evaluate the outer function using the answer given. This final answer is the value for the composite function evaluated at the given point. Calculus I 69 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.21.1 Let f(x) = x 2 + 5 and g(x) = 2x + 4. a.) Find (f g)(1). First, evaluate g(1): g(1) = 2(1) + 4 = 6. Then, evaluate f(6): f(6) = (6) 2 + 5 = 36 + 5 = 41. Thus, (f g)(1) = f( g(1) ) = f(6) = 41. b.) Find g composed with f at the value 1. Evaluate f(x) at x = 1 (f(1) = 1 2 + 5 = 6), and then plug the result into g(x). Hence, (g f)(1) = g( f(1) ) = g(6) = 2(6) + 4 = 16. Calculus I 70 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.22 Find a composite function and its domain. Instead of separately evaluating each component, f(x) and g(x), at a specific point, we want to form an equation for the composite function h(x). Notice that h(x) = (f g)(x) = f( g(x) ); so, we can think of this process as substituting the argument g(x) in for every x in f(x). This composition yields a new equation h(x). Note that order is important in function composition, as g f and f g are often not the same. For example, let f(x) = x + 1 and g(x) = 2x + 1. Suppose we want to evaluate the composition of these two functions at x = 1. If we first evaluate g(x), then we have: However, if we evaluate f(x) first, then we have: Therefore, in general, (f g)(x) (g f)(x). (f g)(1) = f( g(1)) = f(3) = 4. (g f)(1) = g( f(1)) = g(2) = 5. To find the domain of the composite function h(x) = (f g)(x), there are several things we must check: a.) First, find the domain of the inner function, denoted D g. Any numbers not in D g cannot be in the domain of the composite function D h. b.) Second, find the domain of the outer function, denoted D f. For the composite function, the domain of the outer function has to match the range of the inner function. So, if there are real numbers that are not in D f, set the values equal to g(x), the inner function, and solve for x. The resulting x-values are also not in D h. c.) Last, check the composite function for any other values that appear to not be in the domain. d.) The intersection of the above three pieces of information give D h. Calculus I 71 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.22.1 Let f(x) = 3 x 1 and g(x) = 2 x. Find a.) h(x) = (f g)(x). Substitute every x in f(x) with the expression for g(x): h(x) = (f g)(x) = f (g(x)) ( ) 2 = f x 3 = ) 1 ( 2 x = 3x 2 x b.) The domain of h(x). The domain of the inner function is D g = x R x 0; hence, 0 is not in the domain of the composite function h(x). function is The domain of the outer D f = x R x 1. Thus, we need to know when g(x) = 1, so that we know the values of x that would result in an undefined solution: g(x) = 1 2 x = 1 2 = x Now we know that neither 0 nor 2 are in D h. Finally, we check the finalized equation to make sure there are no additional problem spots. In (f g)(x), it appears that the only problems spot is when x = 2. Therefore, D h = x R x 0, 2. Calculus I 72 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.23 Find the components of a composite function. The previous sections have shown how to find a composite function given two pieces. This section discusses how to break apart composite functions into inner and outer pieces. Looking for things like radicals, parentheses, or exponents often give an idea of where to start separating the function. If you break composite function into its components, you can always check your answer by going backward: find the composition of the two smaller functions, which should be the original composite function. Calculus I 73 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.23.1 Find functions f and g such that h(x) = f(g(x)) for each of the following functions: a.) h(x) = x 3 + 5. Note that there is a square root in h(x); recall, square roots, radicals, parentheses, and exponents are helpful for deciding how to begin separating the function. So, let the inner function g(x) be everything on the inside of the square root, i.e. g(x) = x 3 + 5. The only other part of h(x) that needs to be dealt with is the square root; hence, the outer function f(x) = x. If we were to check, we would see that the composition of the functions gives us back h(x). b.) j(x) = 1 2x + 3. The fraction suggests that we could take our inner function g(x) to be the denominator and the outer function to be f(x) = 1 x. Hence, g(x) = 2x + 3. Again, if we compose these two functions, we obtain j(x). Calculus I 74 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.24 Determine whether a function is one-to-one by using the horizontal line test. This module talks about inverse functions and some of their properties. One of the first things that we want to look at is how to tell if a function is one-to-one, which is a necessary condition for it to have an inverse. We know that for an equation to describe a function, it has to pass the vertical line test - namely, that for every x-value, there is a unique y-value assigned to it. To determine whether or not a function in invertible (saying that a function has an inverse), we want to check if every y-value has a unique x-value assigned to it. To do this, we can look at the horizontal line test. To use the horizontal line test, the function must cross the horizontal line only once on its domain (careful here - sometimes functions are one-to-one on portions, but not all, of their domain). This means that there is a unique x-value for every y-value and a unique y-value for every x-value. Another way to look at this is to consider the technique of composition of functions that we learned in Module 10: A function f(x) has an inverse g(x) if f( g(x) ) = g( f(x) ). In this case, we say that g(x) is the inverse of f(x) and f(x) is the inverse of g(x). We can write this as f 1 (x) = g(x) and g 1 (x) = f(x). Calculus I 75 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.24.1 Are the following functions 1-1? a.) y = x 2 + 4 We can see from the graph below that if we draw horizontal line at y = 0, then the function crosses it twice, so y is not 1-1. b.) y = 1 4 x3 We can see from the graph below that no matter where we place the horizontal line, we will only cross the graph once. Thus, y is 1-1. Calculus I 76 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.25 Find the inverse of a function when you are given a table of values or a list of ordered pairs. If you have a function f(x) with an inverse f 1 (x), then for any point (a, b) that is on f(x), the point (b, a) is on f 1 (x). You can think of this as switching the values of x and y for the function. If you are given a table of points, finding the inverse function evaluated at a point is easy. Let s do an example to illustrate this. Suppose that f(x) is given by the table x 1 2 3 4 5 f(x) 2 4 6 9 12 If we were asked to find f(2), we would just go to where x = 2 and find the f(x)-value associated with this. Using inverse functions is also just this easy. If we were asked to find f 1 (9), insteading of going where x = 9, we go where f(x) = 9. Then f 1 (9) is the x-value associated with f(x) = 9. Therefore, f 1 (9) = 4. Calculus I 77 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.25.1 Given the function f below, find f 1 (3). x 2 1 7 12 f(x) 5 3 5 10 To find the inverse function evaluated at 3, we find where f(x) = 3 and look for the corresponding x-value. So, f(x) = 3 when x = 1, so f 1 (3) = 1. Calculus I 78 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.26 Sketch the graph of f 1 given the graph of f. We know from the previous section that for any point (a, b) on the graph of f(x), the point (b, a) is on the graph of f 1 (x). Given the graph of f(x), we could go through the tedious process of taking every point and flipping the coordinates around, or we could take a shortcut. The act of flipping the coordinates of values of points on f(x) to get points on f 1 (x) is the same as reflecting these points across the line y = x. This gives us an easier way to find the graph of f 1 (x); we can just take the entire graph of f(x) and reflect it over the line y = x. Calculus I 79 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.26.1 Sketch the graph of f 1 when given the graph of f below: To graph the inverse of f, we first draw the line y = x (shown in dashed green below). Now, we simply reflect the graph of f(x) over this line in order to get the inverse function f 1 (x), shown in blue. Calculus I 80 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.27 Find f 1 (x), given f(x). If we are given the function f(x) and told to algebraically solve for f 1 (x), there are three steps to the process: a.) Switch the places of x and y. b.) Solve for y (if possible). c.) The inverse function f 1 (x) equals this y. Switching the places of x and y acts like reflecting the graph of the function over y = x. By solving for y in terms of x, we are just rewriting the new function in a form that we can understand. Depending on the complexity of the original function f(x), it may not be possible to completely solve for y. However, most of the problems that you will face in this class will be fully solvable. The final step is renaming the function that you just solved for as the inverse function f 1 (x). Calculus I 81 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.27.1 Find the inverse of f(x) = 5 2x. To answer this question, let s follow the steps talked about in the objective statement: y = 5 2x Original equation x = 5 2y Interchange x and y x 2 = 5 2y Start solving for y 2y = 5 x 2 y = 5 x2 2 f 1 (x) = 5 x2 2 Rename for the inverse function Calculus I 82 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Objective 1.28 Find f 1 when f is a domain restricted function. The steps to finding f 1 when f has a restricted domain are very similar to finding an inverse for a function without a restricted domain. The only difference is that at the end, we just have to make sure that the domain of the inverse function agrees with the range of the original function. Why are we interested in the range of the original function? By reflecting the function over the y = x line, we are essentially changing ranges into domains and domains into ranges. Thus, the domain of f(x) becomes the range of f 1 (x) and the range of f(x) becomes the domain of f 1 (x). Calculus I 83 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Example 1.28.1 Find the inverse of the function f(x) = x 4 + 2 on the restricted domain x 0. We already know from Objective 11.3 how to find the inverse of a function graphically. Now, though, we have to keep in mind the domain restriction on the original function. Let s first solve for the inverse function algebraically: y = x 4 + 2 x = y 4 + 2 x 2 = y 4 ± 4 x 2 = y There are two real roots to this equation, but we only consider the positive one, so f 1 (x) = 4 x 2. This is done by looking at the graph of the function and seeing what the inverse function should look like graphically. The remaining portion is to look at the domain restriction on f(x). We know that since we have an even root, we can t have negatives underneath the radical. So, x 2. If we were to look at the range of f(x), we d see that it is y 2, so as we said in the objective statement, the range of f(x) becomes the domain of f 1 (x). Calculus I 84 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Exercises 1.28 1. Is each of the following set of ordered pairs a function? (a) {(1, 2), (2, 3), (3, 4), (4, 4)} (b) {(1, 2), (1, 3), (3, 4)(4, 5)} 2. (a) x -2 4 5 4 y 5 3 4-1 (b) x -2-1 0 1 2 y -4 2 0 2 4 3. 2x 2 + 8x + 2y 2 = 0 4. (x 3) 2 4 + (y+1)2 9 = 1 5. 9x 2 4y 2 = 36 6. x 2 + 4x 4y = 0 Calculus I 85 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster

Exercises 1.1 1. Is each of the following set of ordered pairs a function? (a) {(1, 2), (2, 3), (3, 4), (4, 4)} (b) {(1, 2), (1, 3), (3, 4)(4, 5)} 2. Determine whether each of the following data sets is a function. (a) (b) x -2 4 5 4 y 5 3 4-1 x -2-1 0 1 2 y -4 2 0 2 4 3. For the following equations, decide whether or not y is a function of x. (a) x 2 + y 2 = 16 (b) y = 6x 2 + x 3 (c) y = 3x 5x 2 40 4. The function h(x) = 3 is called the function: (a) constant (b) real (c) identity 5. Let f(x) = 3x 2 + 2x 5. Evaluate the following. Calculus I 1 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 6, 2014

(a) f(0) (b) f( x) (c) f(x) (d) f(x + h) 6. Let g(a) = a 5 + a 2 + a 3. Evaluate the following. (a) g(0) (b) g( x) (c) g(2) (d) g(2a) 7. Which function equation corresponds to the following data table? x 0 1 2 3 y 4 0 4 8 8. Given the function f(x) = x 2 + 5x 1, find and simplify the difference quotient. 9. Given the function g(x) = x 2 + (x), find and simplify the difference quotient. 10. Find the domain of h(x) = 2 x + 5 11. Which of the following is the domain for f(x) = x 7? Calculus I 2 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 6, 2014

(a) {x R x 7} (b) [7, ) (c) {x R x > 7} (d) (7, ) 12. Let f(x) = x 2 3 and g(x) = x + 1. Find (fg)(3). 13. Given the tables for f and g, find (f g)(2) and (fg)(0). The tables for f and g are as follows: x -1 0 1 2 f(x) 1-2 7 0 x -1 0 1 2 g(x) 4 1-3 7 14. If the domain of f is all real numbers in the interval [0, 3) and the domain of g is all real numbers in the interval (, 0] [1.9, ), find the domain of f g: 15. Given f(x) = x and g(x) = 2x 6. The domain of also be written as [0, ). True/False. ( ) g (x) is {x R x 0}, which can f 16. Given f(x) = x 3 and g(x) = x 2x 6. Find f + g and its domain. 17. Is the following a graph of a function? Calculus I 3 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 6, 2014

18. Given the graph of h below, find the domain and range. 19. Given the graph of g below, determine the domain and range. 20. Match each description below with the graph that best represents it. (a) The weight of a polar bear as a function of time. (b) The cost of iphones as a function of supply (how many are available). (c) The distance traveled by a person on a bike as a function of time. 21. TRUE/FALSE: The following function is increasing on the interval (1,3). Calculus I 4 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 6, 2014

22. Find the local maximum and minimum of the function given below. 23. Find the local maximums and/or minimums of the function given below. 24. Find f(x) for the following basic function. 25. Which basic graph do each of the following graphs most closely resemble? 26. Sketch the graph of the following function: 3, x < 1 f(x) = x, x 1 Calculus I 5 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 6, 2014

27. Sketch the graph of the following function: 2x + 3, x < 2 h(x) = x + 1, 2 x < 2 x 2, x 2 28. Let (a) Evaluate g(1) (b) Evaluate g(5) g(x) = x 1, x 0 1, 0 < x 1 x 2 1, x > 1 Calculus I 6 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 6, 2014

(d) (e) (f) (g) 29. Let h(x) = (x + 2) 2 1, x < 3 47, x = 3 x, 3 < x < 5 2x + 19, x 5 (a) Evaluate the function h(x) when x = 5 (b) Evaluate the function h(x) when x = 3 (c) Evaluate the function h(x) when x = 5 Calculus I 7 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 6, 2014

30. Given the following sketch of the function g, find g(x). 31. Given the following graph of the function h, find h(x). 32. Write a piecewise defined function that models the following information about postal rates on first class mail. If the letter weighs less than or equal to 1 ounce, it costs 42 cents to mail. If the letter weighs more than 1 ounce but less than or equal to 2 ounces, it cost 67 cents to mail. If the letter weighs more than 2 ounces but less than or equal to 3 ounces, it cost 86 cents to mail. If the letter weighs more than 3 ounces but less than or equal to 4 ounces, it cost $1.08 to mail. Write a function C(x) for the cost depending on the ounces, x. Sketch the graph of the function over the domain described. 33. Write a piecewise defined function modeling the following information about Jessica s summer job selling sno-cones. If Jessica sells under 10 sno-cones during her shift, she only makes $20.00 for that day s work. If she sells between 10 and 20 sno cones, she makes $30.00 for the day. If she sells more than 20 sno-cones but less than 40, she earns $40.00. And if she sells at least 40 sno-cones, she Calculus I 8 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 6, 2014

earns $50.00 for that day. Write a function S(x) for the pay depending on the number of sno-cones sold, x. Sketch the graph of the function over the domain described. 34. Sketch the graph of g(x) = x 3 by first identifying its parent function and then by deciding which translation was used. 35. Sketch the graph of y = f( x) + 2 where the parent function f(x) = x. 36. Given the graph of f below, sketch the graph of y = f(x 1) + 3. 37. If the point (1, 3) is a point on the graph of f(x), find a point on the graph of y = f ( 1 2 x) 5. 38. Sketch the graph of y = x + 4 + 3. 39. Given the graph of f below, sketch the graph of y = f(2x) + 1 by translating the labelled points as described above. 40. Look at the sketch of the following function. What is its basic shape? How has the basic shape been translated? Determine a formula for this function. Calculus I 9 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 6, 2014

41. Given the function f(x) = e x, find and sketch the graph of the function g(x) obtained by performing the following translations: horizontal shift right 2 units, reflection about the x- axis, and vertical shift down 1 unit. 42. Look at the sketch of the following function. What is its basic shape? How has the basic shape been translated? Determine a formula for this function. 43. The table for f is as follows Calculus I 10 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 6, 2014

x 1 3 4 6 f(x) 0 1 5 9 The table for g is as follows x 0 1 3 9 g(x) 4 6 9 10 (a) Find g composed with f at 3, i.e. (g f)(3) (b) Find f composed with g at 0, i.e. (f g)(0) (c) Find (g f)(1) (d) Find (g g)(3) 44. Let f(x) = x 3 x and g(x) = 4x + 1. (a) Find f composed with g for x = 1. (b) Find (g g)(2). (c) Find (g f)( 1). 45. Let f(x) = x 2 16 and g(x) = x. Find (a) (g f)(x) (b) The domain of (g f) (c) (f g)(x) (d) The domain of (f g) 46. Let f(x) = x x 1 4 and g(x) = x+2. Find (f g)(x) and its domain. 47. For the following functions, find the components f and g: 4 (a) h(x) = (f g)(x) = x 2 + 2x 7 (b) j(x) = (f g)(x) = 2(x + 3) 4 (c) k(x) = (g f)(x) = 5 (x + 1)(x 7) Calculus I 11 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 6, 2014

48. Is the following function one to one? 49. Given the function f below, find f 1 (4). x 3 1 4 4 8 f(x) 4 0 2 4 50. Sketch the graph of f 1 when given the graph of f below: 51. Find the inverse of f(x) = 5 2x x + 3. 52. (TRUE/FALSE) Let f(x) = x 2 + 3 where x 0. Then f 1 (x) = x 3. Calculus I 12 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 6, 2014

Section2 Mathematical Models: Lines and Polynomials Introduction After completing this module, the learner will be able to: Objective 2.1 Write the equation of a line Objective 2.2 Sketch the graph of a line Objective 2.3 Define a polynomial function. Objective 2.4 Identify the graph of a polynomial function. Objective 2.5 Identify the degree of a polynomial function. Objective 2.6 Identify the power function that models the end behavior of a polynomial. Objective 2.7 Find the real zeros and their multiplicities of polynomials when the polynomial is given as a product of linear and quadratic factors. Objective 2.8 Find a formula for a polynomial given its zeros and their multiplicities. Objective 2.9 Determine if the graph crosses or touches the x-axis at a given zero. Objective 2.10 Find a formula for a polynomial when its graph is given. Objective 2.11 Write the equation of a quadratic function in graphing form. Objective 2.12 Find the vertex of a quadratic function. Objective 2.13 Graph a quadratic function. Objective 2.14 Find a formula for the graph of a quadratic function. Calculus I 1 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.15 Find the maximum or minimum value of a quadratic function. Objective 2.16 Solve word problems by using quadratic models. Objective 2.17 Recognize a rational function. Objective 2.18 Find the domain of a rational function. Objective 2.19 Find the x-intercept(s) and y-intercept of a rational function. Objective 2.20 Find the vertical asymptote(s) of a rational function. Objective 2.21 Find the horizontal asymptote of a rational function. Objective 2.22 Find the oblique asymptote of a rational function. Objective 2.23 Find any holes in the graph of a rational function. Objective 2.24 Graph a rational function. Objective 2.25 Given the graph of a rational function, find f(x). Calculus I 2 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.1 Write the equation of a line Example 2.1.1 Find the equation of the line with slope 2 3 and containing the point (1, 2) in the form Ax+By+C = 0. Use point slope form to start: Thus, A = 2, B = 3, and C = 4. Example 2.1.2 y 2 = 2 (x 1) 3 3(y 2) = 2(x 1) 3y 6 = 2x 2 2x + 3y 4 = 0 Find the equation of the perpendicular bisector given the two points (4, 5) and (2, 3). The slope of the line is given by m = 5 ( 3) 4 2 = 4 Thus, the slope of the perpendicular line is given by m = 1 4. The midpoint of the line between the points is given by ( 4 + 2 p = 2, 5 + ( 3) ) = (3, 1) 2 Therefore, we can use point-slope form to find the equation of the line with slope m and point p: y 1 = 1 4 (x 3) y = 1 4 x + 1 4 Objective 2.2 Sketch the graph of a line Objective 2.3 Define a polynomial function. Polynomials are special types of functions of the form p(x) = a n x n + a n 1 x n 1 +... + a 1 x + a 0, where each coefficient a i is a real number and each power on x is a non-negative integer. We are already familiar with functions for which n = 0 (constant functions), n = 1 (linear functions), and n = 2 (quadratic functions). Calculus I 3 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.3.1 Which of the following are polynomial functions? a.) p(x) = 2x 3 + x 1/2 This is not a polynomial, because the exponent of x 1/2 is not an integer. b.) p(x) = 3x 2 This is not a polynomial, because the power on x is negative. c.) p(x) = 5 This is a polynomial since 5 = 5x 0 (recalling that anything to the zero power is equal to one). d.) p(x) = 5x 4 πx + 3 This is a polynomial. Calculus I 4 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.4 Identify the graph of a polynomial function. The graph of a polynomial is smooth and continuous. Continuous means that there are no holes or jumps in the graph, i.e. that you can draw the graph without picking up your pencil. Smooth means that there are no cusps or corners. These properties become important in calculus for finding out various properties of polynomial functions. Calculus I 5 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.4.1 Is the following graph a polynomial? Explain your answer. There is a jump discontinuity at x = 0 (this means that there is a hole in the graph and that the graph can t be made continuous by simply filling in the hole). Therefore, the graph is not a polynomial. Calculus I 6 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.5 Identify the degree of a polynomial function. The degree of a polynomial is determined by the term with the highest power. Generally, polynomials are written in the form p(x) = a n x n + a n 1 x n 1 +... + a 1 x + a 0 where n is the highest power and the a s are coefficients for each x-term. We call a n the leading coefficient. For example, the degree of the function p(x) = 3x 5 x 2 is 5 and the leading coefficient is 3. Calculus I 7 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.5.1 State the degree of the following polynomials: a.) f(x) = 4x 6 + 5x 2 + 4 + 3x 8 The degree of f(x) is 8. Notice that the polynomial isn t written in standard form, but the degree is still the largest power. b.) g(x) = 4x 3 + 5x + 2 The degree of g(x) is 3, and the polynomial is written in standard form. c.) h(x) = 3x 4 (x 2 + 2x 1) 5 At first glance, it may appear that the degree of the polynomial is equal to 4 or to 5, but this is not the case. We have to consider all of the multiplication that may increase the power of an x-term. The highest power in the term (x 2 + 2x 1) is 2; hence, when we apply the power of 5 to this term, the highest degree will be (x 2 ) 5 = x 1 0. Finally, we multiply the first term to our current highest powered term: (3x 4 )(x 1 0) = 3x 1 4. Therefore, the degree of the polynomial is 14. Calculus I 8 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.5.2 Given the polynomial, g(x) = 9x 5 + 4x 6 + 3x, a.) State the degree. The degree of the polynomial is the largest power of x; so, the degree of g(x) is 6. b.) State the leading coefficient. The leading coefficient is 4. Calculus I 9 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.6 Identify the power function that models the end behavior of a polynomial. A power function of degree n has the form f(x) = ax n where a R (a real number) and n Z + (a positive integer). This is called a power function, because the ends of the polynomial f(x) = a n x n + a n 1 x n 1 +... + a 1 x + a 0 will have the same general shape as the power function y = a n x n. As the x values get very large both positively and negatively, the term with the largest degree dominates over all the other terms. Therefore, the power function provides an idea of the overall shape of the graph. If the polynomial is not in standard form, it may be very difficult to multiply out each set of terms just to find the coefficient on the largest degree term. Suppose the function is of the form f(x) = (a 1 x + b 1 ) c 1 (a 2 x + b 2 ) c2 (a n x + b n ) cn. If we multiplied this function out, it would get messy very quickly. Because the power function deals only with large (± ) values of x, we only have to work with the x terms of largest degree. Hence, we have the function P (x) = (a 1 x) c1 (a n x) cn. Now, we can use rules of exponents to distribute the c terms into the parentheses and to combine the exponents on the x terms: P (x) = (a 1 x) c1 (a n x) cn = a c 1 1 xc1 a cn n x cn = ( a c 1 1 ac 2 2 ) acn n x c 1 +c 2 + +c n which is the power function of f. Calculus I 10 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.6.1 Describe the end behavior of the polynomial p(x) = 3x 8 + 6x 5 8x + 1. The end behavior of a polynomial is dictated by its associated power function. In this case, the power function is y = 3x 8. Since the degree of x is even, either both ends will point up or both ends point down. The coefficient of x 8 is negative; hence, both ends will point down as x goes to ±. The following graph illustrates this power function: Even though the graph jostles around in the middle, we are only interested in the end behavior. Calculus I 11 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.6.2 Predict the end behavior of the following functions: a.) f(x) = 5x 4 + 3(x 1) 2 + 2 The power function is y = 5x 4, which is an even function with a positive coefficient. Thus, the graph will have both ends pointing upward: As x, f(x). As x, f(x). b.) g(x) = 2x 7 + 5x 5 + 3x 4 + 1 The power function for this graph is y = 2x 7. This is an odd function; so, one end of the graph will be pointing up and the other pointing down. The leading coefficient determines which end is up and which is down. Since the coefficient is negative, for large negative values of x, y will be positive, and for large positive values of x, y will be negative. Therefore, the left tail of the graph will be pointing upward and the right tail, downward. Thus, we have the following conclusions: As x, g(x). As x, g(x). Calculus I 12 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.7 Find the real zeros and their multiplicities of polynomials when the polynomial is given as a product of linear and quadratic factors. The goal of this objective is to demonstrate how to find zeros of polynomials as well as their associated multiplicities. Every polynomial p(x) = a n x n + a n 1 x n 1 +... + a 1 x + a 0 can be factored into the form p(x) = a(x c 1 ) m 1 (x c 2 ) m 2...(x c s ) ms where c 1, c 2,..., c s are roots (or zeros) of the polynomial, i.e. p(c i ) = 0. For each root c i, we have its associated multiplicity m i. In general, polynomials can only be completely factored if we include complex numbers. However, for the purposes of this course, the polynomials will be factorable over real numbers. Calculus I 13 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.7.1 State the zeros and their multiplicities for the following polynomial. Also state the degree of the factored polynomial. f(x) = 4(x 3) 5 (x + 4) 3 (x 1) The polynomial f(x) has three zeros: x = 3, x = 4, and x = 1. The multiplicities of each zero correspond to the power of its term; hence, x = 3 has multiplicity 5, x = 4 has multiplicity 3, and x = 1 has multiplicity 1. The degree of f(x) is the sum of the multiplicities; so, f(x) has degree 9. Calculus I 14 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.7.2 Given the polynomial, f(x) = 4(x + 5) 3 (x 3) 2 (x 2 + 9) a.) State the degree of the function and the sign of its leading coefficient. The degree of the function is 7. Notice that the 4 is the only negative coefficient of x; hence, the sign of the leading coefficient is negative. b.) List the real zeros and their multiplicities. The real zeros are x = 5 and x = 3. Note that x 2 + 9 has no real zeros (you can use the quadratic equation to check). The multiplicity of x = 5 is 3 and the multiplicity of x = 3 is 2. Calculus I 15 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.8 Find a formula for a polynomial given its zeros and their multiplicities. The goal of this objective is to show how to find a formula for a function when given the zeros of the function and their multiplicities. For example, given the following table of zeros and their multiplicities, we can write the associated function: zeros multiplicity c 1 m 1 c 2 m 2 c 3 m 3 The function can be written in the form f(x) = (x c 1 ) m 1 (x c 2 ) m 2 (x c 3 ) m 3. Calculus I 16 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.8.1 Find the formula for the polynomial with leading coefficient 5 and the following zeros and multiplicities: zeros p(x) = a(x c 1 ) m 1 (x c 2 ) m 2 (x c 3 ) m 3 = 5(x 0) 3 (x ( 2)) 4 (x 7) 1 = 5(x) 3 (x + 2) 4 (x 7) multiplicity 0 3 2 4 7 1 Calculus I 17 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.9 Determine if the graph crosses or touches the x-axis at a given zero The goal of this objective is to explain how to determine when the graph touches or crosses the x-axis, given the zeros of a function and their multiplicities. First, the polynomial needs to be in factored form (i.e. p(x) = a(x c 1 ) m 1 (x c 2 ) m 2...(x c n ) mn ). There are two ways to determine if the graph crosses or touches the x-axis at each zero. The first way is to make a sign chart, which is a simple chart that breaks the x-axis into regions with each zero value as a divider. A generalized sign chart is depicted below: The sign of the function can be found by choosing and evaluating a value in-between each zero. This will tell you if the function is positive or negative on that region. If two consecutive regions are both positive or both negative, then the function only touches the x-axis at that point. However, if the sign changes from negative to positive, or positive to negative, then the function crosses the x-axis at that point. The second way to determine whether a function touches or crosses the x-axis is by looking at the multiplicities of each zero. If the exponent is even, then the graph only touches the x-axis. If the exponent is odd, then the function will cross the x-axis. Calculus I 18 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.9.1 List the real zeros and their multiplicities for p(x) = 2(x 5) 3 (x + 1). Sketch the graph of p(x). The real zeros of p(x) are x = 5 and x = 1. By looking at the exponents of each term, we can find the multiplicities for each zero. Thus, the zero x = 5 has multiplicity 3, and the zero x = 1 has multiplicity 1. We also know that the power function is y = 2x 4, which has both ends pointing upward (since the coefficient is positive). Because the multiplicities are all odd, we know that the graph will cross at each zero. We can also see this in a sign chart: If we plug in values on either side of the zeros, we see that the function is either negative or positive, as shown in the sign chart. The y-intercept is found by evaluating x = 0. These intuitions give a good idea of the graph of p(x), shown below. Calculus I 19 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.9.2 Given the polynomial function f(x) = 4(x + 5) 3 (x 3) 2 (x 2 + 9) a.) Determine whether the graph crosses or touches the x-axis at the zeros. Since the multiplicity of x = 5 is odd, the graph crosses the x-axis. The multiplicity of x = 3 is even, so the graph only touches the x-axis at that point. b.) Find the power function that the graph of f resembles as x ±. The power function that describes this function is p(x) = 4x 7. Calculus I 20 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.10 Find a formula for a polynomial when its graph is given When given a graph, how can you find the associated polynomial of least degree? First, find the zeros of the polynomial. At every zero, note whether or not the function crosses or just touches the x-axis. Also note the direction of the ends of the function - this will serve as a check that your exponents are correct. If both ends of the function point the same direction, then the function should have a even degree; otherwise, the function will have odd degree. When finding a polynomial of least degree, the multiplicities of the zeros should be either one or two, depending on how they affect the graph of the function. Recall that polynomials can be written in the form p(x) = a(x c 1 ) m 1 (x c 2 ) m 2...(x c s ) mt. Hence, the last thing that we need to do to find an associated polynomial for a graph is find the leading coefficient a. However, a point ( x, p(x) ) on the graph is needed for this step. Then, we can substitute values for x and p(x) to solve for a. Calculus I 21 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.10.1 Find the polynomial of least degree for the following graph: First, find the zeros of the function. Then, find the multiplicities of the function. Finally, solve for the leading coefficient a. The first zero is located at x = 3. Since the function crosses the x-axis, the multiplicity is odd. Because we want the polynomial of least degree, let the multiplicity be 1. The second zero is at x = 0 and has even multiplicity; so let the multiplicity equal 2 (the smallest even number). The third zero is at x = 3 with odd multiplicity. Thus, we can write p(x) as p(x) = a(x + 3) 1 (x 0) 2 (x 3) 1 = a(x + 3)x 2 (x 3) To find the value of a, plug in one of the points, say (4, 1), into the equation for p(x): p(x) = a(x + 3)x 2 (x 3) 1 = a(4 + 3)(4) 2 (4 3) 1 112 Therefore, p(x) = 1 112 (x + 3)x2 (x 3). = a(7)(16)(1) = a(112) = a. Calculus I 22 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.26 Write a Quadratic Function In Graphing Form Often, we are given the equation of a parabola in standard form f(x) = ax 2 + bx + c and would like to write it in the form f(x) = a(x h) 2 + k where (h, k) are the coordinates of the vertex (the minimum or maximum point on the parabola). This form is called the vertex form or the graphing form since all of the information needed to sketch the graph of the parabola is readily available. There needs to be a perfect square trinomial (ie, a three-term polynomial that can be factored into a perfect square) in the standard form in order to get the (x h) 2 portion in the graphing form. This is where the term completing the square comes from. To complete the square, we do the following steps: f(x) = ax 2 + bx + c = (ax 2 + bx) + c = a (x 2 + ba ) x+ + c (( = a x 2 + b b )) 2 (( b )) 2 a x + a a a + c 2 2 ( = a x + b ) 2 b2 2a 4a + c ( = a x + b ) 2 b2 2a 4a + 4ac 4a ( = a x + b ) 2 ( b 2 ) + 4ac + 2a 4a Calculus I 23 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.10.2 Complete the square for y = 3x 2 12x 1. y = 3x 2 12x 1 = 3(x 2 4x+ ) 1 = 3 ( x 2 4x + = 3(x 2 4x + 4) 12 1 ( ) ) ( 4 2 ( ) ) 4 2 3 1 2 2 = 3(x 2) 2 13 Calculus I 24 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.11 Find the vertex of a quadratic function. Graphing Form: If the quadratic equation is in the form y = a(x h) 2 + k, then the vertex of the parabola is located at (h, k). Standard Form: If the quadratic equation is of the form y = ax 2 + bx + c, there are two ways to find the vertex of the parabola: a.) Complete the square to get the equation in graphing form b.) Using the fact that the x-coordinate of the vertex is at then k = a h = b 2a, ( ) b 2 ( ) b + b + c. 2a 2a This is obtained by substituting in h for x and solving. Calculus I 25 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.11.1 Find the vertex of f(x) = 3x 2 2x + 1. is We know that this equation is in standard form; hence, the x-coordinate of the vertex h = b 2a = ( 2) = 1 2(3) 3. Now we can substitute h in for x into our equation f(x) and solve for the y-coordinate of the vertex. So, ( ) ( ) 1 1 2 ( ) 1 k = f(h) = f = 3 2 + 1 = 2 3 3 3 3. ( 1 Therefore, the vertex is at the point 3, 2 ). 3 Calculus I 26 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.12, Part A Translations of basic functions. In order to graph a quadratic function by using translations, the function must be in graphing form, f(x) = a(x h) 2 + k. We can define a new function, g(x) = x 2, to use as a parent function. Now we can think of our quadratic function as y = a g(x h)+k. Thus, the vertex of the standard parabola y = x 2 is moved to the point (h, k) and the value of a determines whether the graph has been reflected about the x-axis as well as indicating the vertical stretch or compression factor. To illustrate the ideas presented, let s consider the function f(x) = 2(x + 1) 2. The parent function is g(x) = x 2. Therefore, we can think of f(x) as f(x) = 2g(x + 1). To sketch the graph of f(x), we shift the graph of the basic parabola g(x) = x 2 to the left one unit, then reflect about the x-axis, and finally, stretch the graph vertically by a factor of 2. Calculus I 27 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.12, Part B Using the vertex, intercepts, and orientation. The second method for graphing a parabola is to find the vertex and determine which way the parabola opens. Once you have these two pieces of information, you can make a sketch of the graph. To determine how wide the graph is, plot at least one point near the vertex. For example, let f(x) = x 2 + 4x + 5. The vertex can be found using methods from the previous objective. We have h = 4 2( 1) = 2 and k = f(2) = (2)2 + 4(2) + 5 = 9. Therefore, the vertex is (2, 9). Since a = 1, the graph of the parabola opens down. To find the coordinates of another point on the graph, choose an x-value near the vertex. In our example, let x = 3. To find the y-coordinate, evaluate f(3): f(3) = (3) 2 + 4(3) + 5 = 8. Therefore, the point (3, 8) is also on the graph of f(x). It is also useful to find the x-intercepts, because in many cases the graph will cross the x-axis. We will discuss how to find x-intercepts in the next section, and we will give a graph of f(x) = x 2 + 4x + 5. Calculus I 28 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.12, Part C Finding the roots of the parabola. To find the x-intercepts (or roots) of the parabola, f(x), we need to solve the equation f(x) = 0. In the previous section, we considered the function Let s find the x-intercepts of f(x). First, we set f(x) = 0 and factor the equation: f(x) = x 2 + 4x + 5. 0 = f(x) = x 2 + 4x + 5 = (x 2 4x 5) = (x 5)(x + 1) Notice that at least one piece of the factored equation has to be zero. So, either x 5 = 0 or x + 1 = 0. Hence, if f(x) = 0, then either x = 5 or x = 1. Therefore, the x-intercepts are ( 1, 0) and (5, 0). The following is a graph of f(x) = x 2 + 4x + 5: Calculus I 29 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.12.1 Sketch the graph of g(x) = 2(x 4) 2 + 8. Identify the vertex, domain, range, y-intercept, and whether the graph opens up or down. Recalling that our parent function for g(x) is x 2, we can use our translation properties to draw the following graph: a.) To find the vertex, we can just read the values of h and k from g(x), because the function is in graphing form. Therefore, the vertex is located at the point (4, 8). b.) The domain of g(x) is the same as the domain of the parent function x 2 ; so the domain is all reals R, or (, ). c.) Since a is positive (recall f(x) = a(x h) 2 + k), the graph opens upward. Hence, the vertex is the minimum point on the graph and the y-coordinate is the smallest value in the range. Therefore, the range is [8, ). d.) To find the y-intercept, let x = 0 and solve for y: g(0) = 2(0 4) 2 + 8 = 2( 4) 2 + 8 = 2(16) + 8 = 40 Therefore, the y-intercept is at the point (0, 40). Calculus I 30 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.13 Find the maximum or minimum value of a quadratic function. To determine whether or not a parabola has a minimum or a maximum value, we look at the sign of the coefficient a on the x 2 term. If a is negative, then the parabola will open down and the vertex is the maximum of the graph. If a is positive, the parabola will open upward and the vertex is the minimum of the graph. The maximum/minimum function value is the value of k for the vertex (h, k). Calculus I 31 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.13.1 Find the minimum or the maximum of f(x) = 3x 2 2x + 1. Since a = 3 < 0, the parabola opens downward; hence, the vertex is the maximum point. Recall that there are two ways to find the vertex: a.) We will first find the vertex by completing the square to obtain the graphing form. f(x) = 3x 2 2x + 1 ( ) = 3 x 2 + 2 2 2 3 x + 3 + 1 + 2 ( = 3 x 2 + 2 3 x + 1 ) + 4 9 3 ( = 3 x 2 + 1 ) 2 + 4 3 3, ( ) 2 2 3 2 So, the vertex of the parabola is ( 1 3, 4 3), and hence, the maximum value is 4 3. b.) The second way to find the vertex is by using the formulas h = b 2a and k = f ( b 2a). h = b 2a = 2 ( 3)(2) = 1 3 ( k = 3 1 2 ( 2 3) 1 ) + 1 3 = 4 3, Again, the maximum value of f(x) is 4 3. Calculus I 32 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.14 Find a formula for the graph of a quadratic function When given at least two points,it is possible to find the quadratic equation associated with the parabola. For this class, usually one of the points that you are given will be the vertex, which makes using the graphing form of the quadratic formula easiest to use. The second point can be used to find the value of a for the parabola. Calculus I 33 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.14.1 Find f(x) for the function given below: We are given two different points: (3, 5) and (2, 1) (the vertex). We can use the graphing form of the quadratic equation to find f(x): Plugging in the point (3, 5) yields: f(x) = a(x h) 2 + k = a(x 2) 2 + 1 5 = a(3 2) 2 + 1 = a(1) 2 + 1 = a + 1 4 = a. Therefore, the graphing form of f(x) = 4(x 2) 2 + 1. We can change this into standard form by expanding the squared term and simplifying: f(x) = 4(x 2) 2 + 1 = 4(x 2 4x + 4) + 1 = 4x 2 16x + 16 + 1 = 4x 2 16x + 17, Therefore, the standard form of the parabola is f(x) = 4(x 2) 2 + 1. Calculus I 34 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.15 Solving word problems by using quadratic models When given a story problem, the steps to find the solution are very similar to the steps presented in Module 6. a.) Read the problem twice and underline any important information. b.) Draw a picture to help you visualize the problem. c.) Assign variables to elements of the problem. d.) Create a function to model the objective in the problem. e.) Use this function to answer questions about the objective of the problem. Calculus I 35 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.15.1 A farmer with 2000 meters of fencing wants to enclose a rectangular plot that borders on a river. If the farmer does not fence the side along the river, what is the maximum area that can be enclosed with the 2000 meters of fencing? The first thing that we can do is draw a picture: Let x denote the length of the fence perpendicular to the river. The perimeter of the fenced part of the rectangular plot is given by P = y + 2x; hence, The formula for area is given by A = xy; so P = 2000 = y + 2x y = 2000 2x. A(x) = x(2000 2x) = 2x 2 + 2000x. This equation is logical, because the parabola opens downward, i.e. there is a maximum value for the area. To find the value of the maximum area, we can use the formula h = b 2a to find h and then use this to find k: h = b 2a = 2000 = 500. 4 Thus, k = A(h) = A(500) = 2(500) 2 + 2000(500) = 500, 000. Therefore, the maximum possible area enclosed is 500, 000 square meters. Calculus I 36 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.16 Recognize A Rational Function A rational function, r(x), is a ratio of two polynomials, i.e. r(x) = p(x) q(x), where p(x) and q(x) are polynomials. Recall that a polynomial has the form p(x) = a n x n + a n 1 x n 1 +... + a 1 x + a 0, in which each power is a non-negative integer and each coefficient is a real number. Some examples of rational functions are: a.) r(x) = 1 x b.) r(x) = 2x + 1 3x 2 + 2x + 3 c.) r(x) = 5x3 + 2x + 1 x 3 Calculus I 37 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.16.1 Are the following functions rational functions? a.) f(x) = x + 2 x 2 3 This is a rational function since both x + 2 and x 2 3 are polynomials. b.) g(x) = 2x + 1 x Note that 2x + 1 can be written as (2x + 1) 1/2, but this is still not a polynomial; the exponent is not an integer. Therefore, g(x) is not a rational function. Calculus I 38 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.17 Find the domain of a rational function. The goal for this objective is to demonstrate how to find the domain of a rational function. Recall that for any polynomial, the domain is (, ). However, rational functions are quotients, and so we need to be concerned about the values for which the denominator equals zero. Therefore, the domain of a rational function r(x) = p(x) q(x) is all real numbers, except for the values of x for which q(x) = 0, i.e. x R q(x) 0. Calculus I 39 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.17.1 Find the domain for the following rational functions: a.) f(x) = 1 x 3 We only need to check when the denominator equals zero, i.e. x 3 = 0. Hence, x = 3 gives a zero in the denominator. Therefore, the domain of f(x) is (, 3) (3, ), or {x R x 3}. b.) g(x) = x 3 x 2 5 We need to check when x 2 5 = 0; so, x 2 5 = 0 x 2 = 5 x = ± 5 Therefore, the domain of g(x) is x R x ± 5, or (, 5) ( 5, 5) ( 5, ). c.) h(x) = 5x2 x 2 + 1 When we set the denominator equal to zero, we have x 2 + 1 = 0 x 2 = 1. However, we cannot solve this equation unless we are working in the field of complex numbers. Therefore, the domain of h(x) is all real numbers R, or (, ). Calculus I 40 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.17.2 Find the domain of the rational function f(x) = x(1 x) 3x 2 + 5x 2. The domain of the function f(x) is all real numbers except where the denominator equals zero. Thus, 0 = 3x 2 + 5x 2 = (3x 1)(x + 2) x = 1 3 or x = 2. Therefore, the domain of f(x) is x R x 1 3, 2, or (, 2) ( 2, 1 3 ) ( 1 3, ). Calculus I 41 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.18 Find the x-intercept(s) and y-intercept of a rational functional Finding the x- and y-intercepts of a rational function follows the same steps as finding the intercepts of any other function: to find the x-intercept(s), set y = 0 and solve for x, and to find the y-intercept, set x = 0 and solve for y. Note that not all rational functions have x- and y-intercepts. For example, the function f(x) = 1 x has neither. When x = 0, you get a divide-by-zero case, which means the function is undefined. It is also not possible to get y = 0, because there is no solution to the equation 0 1 x. We will see in later objectives that vertical and horizontal asymptotes sometimes cause intercepts to not exist. Calculus I 42 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.18.1 Find the y- and x-intercept(s) of the following functions: a.) f(x) = x 3 x 2 9 First, let s find the y-intercept. Let x = 0, which yields f(0) = 3 9 = 1 3. Hence, the y-intercept is ( 0, 3) 1. To find the x-intercept(s), set f(x) = 0 and solve for x. Note: the only time a fraction equals zero is when the numerator is zero. Therefore, x 3 = 0 x = 3. So, the x-intercept is (3, 0). b.) g(x) = 2x2 32 x When finding the y-intercept, we set x equal to zero; however, for this function, if we substituted zero in for x, we would have a zero in the denominator. Hence, the y-intercept is undefined. Remember, for the x-intercept, set the numerator equal to zero. This yields 2x 2 32 = 0 x 2 16 = 0 x 2 = 16 x = ±4. Therefore, there are two x-intercepts: the first at ( 4, 0) and the second at (4, 0). Calculus I 43 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.19 Find the vertical asymptote(s) of a rational function This objective is about finding the equations for vertical asymptotes of rational functions. A vertical asymptote is an imaginary vertical line that the function approaches but never touches. Vertical asymptotes occur when the rational function in its simplified form is undefined. For example, the function (x a) r(x) = (x a)(x b) is undefined at x = a and x = b. However, r(x) isn t simplified, because the (x a) terms from the numerator and the denominator can be cancelled. Simplified, we see that r(x) = 1 (x b) is undefined at x = b, and since r(x) is now in simplified form, x = b is the equation of the vertical asymptote. It is important to note that a function can never have a value on a vertical asymptote, in contrast to a horizontal asymptote, discussed in Section 9.5. So what happens to x = a? We know that this isn t part of the domain of r(x), because r(a) is undefined. We will talk about this situation in Section 9.7. Calculus I 44 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.19.1 Find the vertical asymptote of f(x) = 2(x 1)(x + 1) (x 1)(x + 7). The x 1 terms cancel out; so, f(x) is simplified to f(x) = 2(x + 1) (x + 7). Hence, the vertical asymptote is at x = 7. The graph of the function f(x) is below. Notice the blue vertical asymptote at x = 7. We will discuss the reasoning for the open circle at ( 1, 1 2) in section 9.7. Calculus I 45 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.19.2 State the domain and the vertical asymptotes of g(x) = 5 x 2 9. The domain of the function is x R x 3, 3. therefore, the vertical asymptotes are x = 3 and x = 3. The function is in simplified form; Calculus I 46 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.20 Find the horizontal asymptote of a rational function. Horizontal asymptotes describe the end behavior of the graph of a rational function. A function can have real values at a horizontal asymptote, in comparison to that of a vertical asymptote. Note that although there may be more than one vertical asymptote, there can be at most one horizontal asymptote for a rational function. For a function r(x) = p(x) q(x), the equation y = b is a horizontal asymptote if as x approaches ±, r(x) approaches b. In determining whether or not a horizontal asymptote exists, we can look at three cases: a.) If the degree of p(x) is smaller than the degree of q(x), then y = 0 is a horizontal asymptote. b.) If the degree of p(x) equals the degree of q(x), then y = a n b n is a horizontal asymptote, where a n and b n are the leading coefficients of p(x) and q(x), respectively. c.) If the degree of p(x) is greater than the degree of q(x), then there is no horizontal asymptote. Calculus I 47 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.20.1 Find the horizontal asymptote of g(x) = x2 x 3 + 5. The degree of p(x) is 2 and the degree of q(x) is 3. For this function, we will use the first case to fin the horizontal asymptote y = 0. This is verified by the following graph, in which the blue line is the vertical asymptote at x = 3 5 and the green line is the horizontal asymptote. Note that in this example, the function g(x) touches the horizontal asymptote at x = 0. We can see this by solving the equation 0 = x2 x 3 + 5 0 = x2 x = 0 Calculus I 48 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.20.2 Find the horizontal asymptote of f(x) = 3x + 1 x + 2. The degree of both the numerator and the denominator is one; hence, the horizontal asymptote is the ratio of the leading coefficients. Therefore, y = 3 1 = 3 is the equation for the horizontal asympote. Note that the equation for the vertical asymptote is x = 2. This is verified in the following graph, with the vertical asymptote in blue and the horizontal in green. Calculus I 49 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.20.3 For the following function, f(x) = 2x2 x 6 x 2 4 find the horizontal and vertical asymptotes, the domain, y- and x-intercepts, and the following values: a.) The value as x +. b.) The value as x. c.) The value as x 2 +, or as x approaches 2 from the right. d.) The value as x 2 +, or as x approaches 2 from the right. Use these answers to graph f(x). First notice that the degree of the numerator is the same as the degree of the denominator; hence, the horizontal asymptote is the ratio of the leading coefficients: y = 2 1 = 2. To answer the rest of these questions, we need to factor f(x): f(x) = 2x2 x 6 x 2 4 Hence, the domain of f(x) is x R x 2, 2. = (2x + 3)(x 2) (x 2)(x + 2). Now, let s simplify f(x): f(x) = (2x + 3)(x 2) (x 2)(x + 2) = 2x + 3 x + 2. Even though we were able to cancel out the term x 2, the point when x = 2 is undefined (we ll discuss this in section 9.7). From the simplified form of f(x), we know that x = 2 is a vertical asymptote. ( We still ) need to find the x- and y-intercepts. Plugging in x = 0 gives y = 3 2 ; so, the y-intercept is 0, 3 2. Setting the numerator equal to zero yields x = 3 2 ; hence, the x-intercept is ( 3 2, 0). To obtain enough information for graphing the function, we need to determine what happens as x approaches the values listed above. a.) The value as x + is also the end behavior of the right side of the graph, which is described by the horizontal asymptote. Thus, as x +, we know y 2. b.) The value as x is the same as for x +, because we have a horizontal asymptote. So, as x, we know y 2. Calculus I 50 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

c.) The value as x 2 + is the expected y-value of the undefined point at x = 2 in the graph. We can plug x = 2 into the simplified equation for f(x) to obtain this value: f(2) = 2(2) + 3 2 + 2 = 7 4. Hence, as x 2 +, we know y 7 4. d.) The value as x 2 + is how f(x) approaches the vertical asymptote. If we plug in values that are slightly bigger than 2, such as 1.9 or 1.99, we see that the denominator is approaching zero, and that the function is negative. Hence, f(x) as x 2 +. If we were to look at x 2, we would see that the denominator is also approaching 0, but this time the function is positive. Hence, f(x) as x 2. Combining all of this information gives the following graph: Calculus I 51 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.21 Find the oblique asymptote of a rational function. Oblique (or slant) asymptotes are special cases where the degree of the numerator of a rational function is one more than the degree of the denominator. In this case, the equation of the oblique asymptote is given by a linear equation y = mx + b. Note that if a oblique asymptote exists, a horizontal asymptote can t exist, and vice versa. To find the equation of the oblique asymptote, take p(x) and divide it by q(x) (generally by using polynomial long division). The quotient of this operation is called the oblique asymptote. Note that there may be a remainer left after the long division, but this will not affect the function for very large or very small values of x. Calculus I 52 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.21.1 Find the equation for the oblique asymptote of the rational function f(x) = 3x3 1 x 2. The degree of the numerator is exactly one greater than the degree of the denominator; so, there is an oblique asymptote. Using long division, we can find the oblique asymptote: 3x x 2 ) 3x 3 1 3x 3 Since the remainder will not affect the function when the values of x are really small/large, the equation for the oblique asymptote is just y = 3x. The graph of the function f(x) is below: 1 Calculus I 53 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.21.2 For the following function, g(x) = 3x3 6x 2 + x x 2 4 state the domain, the y- and x-intercepts, and any horizontal, oblique, and vertical asymptotes. Also, give the following values: a.) The value as x. b.) The value as x. c.) The value as x 2 +, i.e. as x approaches 2 from the right. d.) The value as x 2 +, i.e. as x approaches 2 from the right. Use these answers to graph g(x). Let s first check to see if the function has a horizontal or oblique asymptote. Remember, for these two types of asymptotes we need to compare the degree of the numerator to the degree of the denominator. For this example, the degree of the numerator is one higher; hence there is an oblique asymptote. Now, we need to do long division: 3x 6 x 2 4 ) 3x 3 6x 2 +x Therefore, the oblique asymptote is y = 3x 6. (3x 3 +0x 2 12x) 6x 2 +13x +0 ( 6x 2 +0x +24) 13x Now, let s factor the function g(x) by using the quadratic formula: 24 g(x) = x(3x2 6x + 1) (x 2)(x + 2) = x(x 1 + 6 3 )(x 1 6 3 ). (x 2)(x + 2) Now we can find the x- and y-intercepts. If we let x = 0, we find the y-intercept at the point (0, 0). Then we can set the numerator equal to zero and solve to find the x-intercepts; hence, the points (0, 0), (1 6 3, 0), and (1 + 6 3, 0). By looking at the denominator, we know the domain of g(x) is {x R x 0, 1 ± And, since nothing cancels out when we simplify g(x), we have two vertical asymptotes: one at x = 2 and the other at x = 2. Finally, we just need to evaluate the following: Calculus I 54 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster 6 3 }. Last update: August 7, 2014

a.) As x, the function g(x) follows along the oblique asymptote, which is a linear function with positive slope; so, g(x). b.) Using the same argument as above, as x, the function g(x). c.) As x 2 +, the function g(x) is approaching the vertical asymptote at x = 2. If we evaluate values slightly larger than 2, then the denominator approaches zero and g(x) is positive; hence, g(x). d.) Using the same argument as above, as x 2 +, the function g(x). Note: by a similar argument, if x 2 or x 2, then g(x). We have enough information to sketch the graph, which is given below, but we need to examine the origin more closely. First, notice that each of the x-intercepts has multiplicity 1. Hence, the graph crosses the x-axis at each x-intercept. We also know that as x 2 +, the function g(x). So, the graph of g(x) slopes downward toward the origin, following along the vertical asymptote at x = 2. Then the function crosses the x-axis at the origin, and begins to curve upward toward the zero, or x-intercept, at the point (1 6 3, 0). After a short while, the graph curves downward again, crosses the x-axis 6 again at the zero (1 + 3, 0), and finally approaches along the vertical asymptote at x = 2. Below is a depiction of the function near the origin: Calculus I 55 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Calculus I 56 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.22 Find any holes in the graph of a rational function. When rational functions are undefined, they either have a vertical asymptote or a hole in the graph at that x-value. Holes are values of x for which f(x) does not exist, i.e. the graph is missing the point (x, f(x)). The process of finding holes is described below: a.) Given a function r(x) = p(x), first factor the numerator and denominator. Recall that a q(x) polynomial is in a factored form when p(x) = a(x c 1 ) m 1 (x c 2 ) m 2...(x c n ) mn. b.) Determine if there are any common terms in the numerator and denominator that would cancel out if simplifying the function r(x). The locations of the holes can be determined by setting these terms equal to zero and solving for x. Recall that any remaining factors in the denominator, when set equal to zero, give the equations for the vertical asymptotes. c.) The corresponding y-values of the hole can be found by evaluating the x-value using the simplified form of r(x). Note: the simplified form of r(x) must be used to evaluate the hole ; otherwise, the y-value is undefined. Calculus I 57 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.22.1 Identify any asymptotes and holes in the graph of g(x) = x 4 x 2 x 12. In the function g(x), the degree of the numerator is one less than the degree of the denominator; so, there is a horizontal asymptote at y = 0. The first step in finding the vertical asymptotes and/or holes in the graph of g(x) is to factor the denominator of g(x): g(x) = = = x 4 x 2 x 12 (x 4) (x + 3)(x 4) 1 x + 3 Notice that there are two places for which the domain of g(x) does not exist: x = 3 and x = 4. However, in the simplified form of g(x), there appears to be only one place where the function does not exist: x = 3. Therefore, there is a hole when x = 4. Using the simplified form of g(x), if we evaluate when x = 4, we get g(4) = 1 7 ; hence, the hole is located at ( 4, 1 7). The remaining equation, x = 3, is the equation for the vertical asymptote. The graph of g(x) is below: Calculus I 58 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.23 Graph a rational function We now have all of the tools that we need to graph a rational function. The following procedure explains how to graph the function r(x) = p(x) q(x) : a.) Write r(x) in its factored form. Use this to find the domain of r(x). b.) Find the common terms (x c 1 ), (x c 2 ),..., (x c n ) that can be cancelled out from the numerator and deominator. The x-values of the holes in the graph are x = c 1, x = c 2,..., x = c n. Then, cancel out these terms to find the simplified form of r(x). c.) Take any remaining terms in the denominator and set them equal to zero; these are the equations for the vertical asymptotes of the graph. Draw these with a vertical dashed line on your sketch. d.) Compare the degree of the numerator to the degree of the denominator: does a horizontal or an oblique asymptote exist? (If you don t remember the conditions on the comparisons of the degrees, go back to Sections 9.5 and 9.6) Draw these equations, if they exist, on the graph with dashed lines as well. e.) Find the y-intercept of the function by letting x = 0. Find the equation for the x-intercepts by setting the numerator equal to zero and solving for x. Plot all of these points. f.) Determine the end behavior of the function by using a sign chart. g.) Now you have a bunch of points and information about the behavior of the graph; so, put it all together for the final sketch. Calculus I 59 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.23.1 Sketch the graph of f(x) = (2x + 1)(x 3) (x 3)(x + 1). Note that f(x) is already in factored form. The domain of f(x) is x R x 3, 1. Before we simplify f(x), note that we can cancel the term (x 3), because it is in both the numerator and the denominator. Hence, there is a hole in the graph at x = 3. Simplifying f(x) yields f(x) = 2x + 1 x + 1. The remaining term in the denominator gives a vertical asymptote at x = 1. Since we have f(x) in simplified form, we can evaluate the function at x = 3 to find the y-value of the hole; thus, the hole is at the point ( 3, 7 4). The next step is to look for any horizontal or oblique asymptotes. Since the degree of the numerator equals the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients; hence, the horizontal asymptote is y = 2. We have now taken care of the domain, the holes, and the asymptotes. The next step is to find the intercepts. Setting x = 0 yields the y-intercept: f(0) = 2(0) + 1 (0) + 1 = 1. Hence, the y-intercept is at (0, 1). For the x-intercept, set the numerator equal to zero. Thus, the x-intercept is x = 1 2. Finally, we can make a sign chart to determine the end behavior of the graph. The key in making a sign chart is to know which points to choose as the dividing points: it is generally easiest to use the x-intercept(s) as the dividing point(s). In this case, let s choose the following values: x f(x) 2 3 0 1 2 1 1 3 Remember that a function can never cross the vertical asymptotes, and that the end behavior follows the horizontal asymptote. The following graph combines all of these pieces. Calculus I 60 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Calculus I 61 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Objective 2.24 Given the graph of a rational function, find f(x) When given a function, we now know all of the steps needed to sketch a graph of the function. We can now ask ourselves, if given a graph, can we determine the associated function? Suppose we are given the graph below, and asked to find its associated function f(x). First, notice that there is a vertical asymptote; hence, the graph is a rational function. Thus, f(x) has the form, f(x) = p(x). Now, using the information we have learned in the past several sections, q(x) here is a procedure for determining the function when given the graph: a.) Find the equations for any holes or vertical asymptotes of the graph. Write these equations in such a way that they will be useful for rewriting f(x) in factored form. In our example, there is a vertical asymptote at x = 1; hence, the term (x + 1) is in our function. There is a hole in the graph at x = 3; thus, the term (x 3) is contained in our function. Because the vertical asymptotes and holes are undefined, we know that x = 3 and x = 1 are not in the domain of f(x). b.) Construct the denominator of f(x). Since x = 3 and x = 1 are not in the domain of f(x), and polynomials have domain equal to R, the terms (x 3) and (x + 1) are in the denominator. Notice that these two terms are the ONLY terms in the denominator; otherwise, we would have more vertical asymptotes or Calculus I 62 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

holes in our graph. Currently, our function can be described as follows: f(x) = p(x) (x 3)(x + 1). c.) Find the equations for any horizontal or oblique asymptotes. Conveniently, in the graph, there is a green dashed line denoting a horizontal asymptote. Using the rules presented in Sections 9.5 and 9.6, we know that the degree of the numerator has to be less than or equal to the degree of the denominator in order to have a horizontal asymptote. Since the asymptote is not at y = 0, the degree of the numerator must be equal to the degree of the denominator. Therefore, both the numerator and the denominator have degree 2. Also, since the equation for the horizontal asymptote is y = 2, the ratio of the leading coefficients is 2. d.) Construct the numerator. This construction has several parts: (a) Add the term corresponding to the hole in the graph to the numerator and determine how many other terms there are based on the degree of the denominator. We know that since x = 3 is a hole in the graph, the term (x 3) must be in the numerator. So, r(x)(x 3) f(x) = (x 3)(x + 1). And, because we know the numerator has degree 2, we know that r(x) has degree 1. Hence, we now have (ax + b)(x 3) f(x) = (x 3)(x + 1). (b) Use the x- and y-intercepts to determine the numerator of f(x) when it is in simplified form. In simplified form, f(x) = ax+b x+1. Given the information about the horizontal asymptote, we know that the ratio of the leading terms is 2; hence, a = 2. The y-intercept (at the point (0, 1)) gives 1 = f(0) = b 1 b = 1. Therefore, f(x) = 2x + 1 x + 1. To find the x-intercept, set the numerator equal to 0: 2x + 1 = 0 2x = 1 x = 1 2. Because the graph determines that the x-intercept is at the point ( 1/2, 0), we know that our values of a and b are correct. Calculus I 63 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

e.) Use a point to check that the function f(x) in its unsimplified form works. (2x + 1)(x 3) The final function is f(x) = (which was also the function in Example 9.8.1, (x 3)(x + 1) in case you want to check the steps going the other way). Looking at the graph, we can find the point (-2,3). So, if evaluated, f( 2) = 3. Let s check: f( 2) = ( 3)( 5) ( 5)( 1) = 3 1 = 3. These are the steps that you follow to find the equation for the function given its graph. Calculus I 64 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Example 2.24.1 Find a function f(x) to fit a curve with asymptotes at y = 1, x = 2 and x = 2. There are two vertical asymptotes (x = 2 and x = 2) and one horizontal asymptote (y = 1). Therefore, the denominator of the function is (x + 2)(x 2). The horizontal asymptote is not y = 0; so, the degree of the numerator is equal to the degree of the denominator. We also know that the ration of leading coefficients is 1. Keeping it simple, let the numerator equal x 2, so that one option for f(x) is: x 2 f(x) = (x + 2)(x 2). Another option for f(x) is f 2 (x) = x2 x 1 (x + 2)(x 2), because the degree of the numerator is still 2 and the function has leading coefficient 1. A third possible equation for the function is f 3 (x) = 2x 2 + 3 2(x + 2)(x 2). Calculus I 65 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 7, 2014

Exercises 1.2 1. True/ False: The function p(x) = 3x 2 + x 1 is a polynomial function. 2. True/False. The following is the graph of a polynomial function. 3. True/False. The following graph is a polynomial function. 4. What is the degree of each of the following polynomials? p(x) = 4(x 1)(x 2 + 2) 3 q(x) = 15(x + 2)(x 1)(x 4 + 2x + 9) 6 5. Given the polynomial, p(x) = πx 2 6 + 3x 1 5 x 7 + x 17, find the degree and the leading coefficient. Calculus I 1 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 6, 2014

6. Given the function f(x) = 2x 3 + 3x + 1, state the power function and describe the end behavior. 7. Predict the end behavior of each of the following graphs: (a) p(x) = 3x 6 + x 5 + 2x 3 x 2 + 4x 13 (b) q(x) = 7x 7 + 6x 6 5x 5 + 4x 4 3x 3 + 2x 2 + x 8. True/False: The polynomial p(x) = (x + 3) 4 (x 1) 2 has a zero of 3 with multiplicity 4. 9. Given the polynomial h(x) = 3(x 2) 6 (x + 7) 3 ( 14)(x 3 1) 2, give the degree of the function, the sign of the leading coefficient, and the zeros and their multiplicities. 10. Find the formula for a polynomial with leading coefficient 7 and the following zeroes and multiplicities: zero multiplicity 2 1 0 2-3 4 11. True/False. The graph of p(x) = x 3 (x 1) 4 crosses the x-axis at x = 1 12. Given the polynomial p(x) = 87(x 2 4)(x + 13) 5 (x 7) 3 (a) Determine whether the graph crosses or touches the x-axis at the zeros. (b) Find the power function that the graph of p(x) resembles as x ±. 13. Given the polynomial function f(x) = 4(x + 5) 3 (x 3) 2 (x 2 + 9) Calculus I 2 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 6, 2014

(a) Determine whether the graph crosses or touches the x-axis at the zeros. (b) Find the power function that the graph of f resembles as x ±. 14. Given the polynomial p(x) = 87(x 2 4)(x + 13) 5 (x 7) 3 (a) Determine whether the graph crosses or touches the x-axis at the zeros. (b) Find the power function that the graph of p(x) resembles as x ±. 15. Given the polynomial, g(x) = 3(x 2) 2 (8x + 4) 6 (x 2 9) (a) Determine whether the graph crosses or touches the x-axis at the x-intercepts. (b) Find the power function that the graph of g resembles as x ±. 16. Which of the following functions could be the polynomial of least degree for the graph given below? (a) p(x) = (x + 2)(x 3) 2 (b) p(x) = (x + 2)(x 3) 2 (c) p(x) = (x + 2) 2 (x 3) Calculus I 3 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 6, 2014

17. Find the polynomial of least degree for the following graph: 18. Complete the square on the function y = 5x 2 + 20x 1 to get it in graphing form. 19. Complete the square on the function f(x) = 3x 2 9x + 13. 20. Find the vertex for each of the following: (a) j(x) = 4(x 3) 2 + 2 (b) k(x) = x 2 + 6x + 1 21. Sketch the graph of g(x) = (x + 3) 2 1. Identify the vertex, domain, range, y-intercept, and whether the parabola opens up or down. 22. Sketch the graph of h(x) = 1 2 x2 x 4. Identify the vertex, domain, range, x-intercepts, y-intercept, and whether the parabola opens up or down. 23. Sketch the graph of h(x) = 1 2 x2 x 4. Identify the vertex, domain, range, x-intercepts, y-intercept, and whether the parabola opens up or down. 24. Find the maximum or minimum for each of the following: (a) j(x) = 4(x 3) 2 + 2 (b) k(x) = x 2 + 6x + 1 (c) r(x) = 3x 2 12x + 22 25. Given the following sketch of a graph, find the function. Note: the vertex is at (4, 6). Calculus I 4 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 6, 2014

26. Given the following sketch of a parabola with vertex (7, 3), find the function. 27. A farmer with 4000 meters of fencing wants to enclose a rectangular plot that borders on a river. If the farmer does not fence the side beside the river, what is the maximum area that can be enclosed with the 4000 meters of fencing? 28. A zoo is building a new exhibit for prairie dogs. They have 1000 feet of fencing available and want to make the exhibit as large as possible. What is the maximum area that can be enclosed using the available fencing? 29. Are the functions below rational functions? Why or why not? (a) h(x) = 4x3 2x 3 x + 5 (b) g(x) = πx2 3 x 4 2x + 14 (c) r(x) = x2 4 (x + 1) 1 (d) p(x) = x 1 1 Calculus I 5 c 2014-16 Brenda Burns-Williams and Elizabeth Dempster Last update: August 6, 2014