Pre Calculus 1 Polynomial and Rational Functions Day 1: Quadratic Functions Guiding question: How are quadratic functions related to polynomial functions? Students will begin by writing everything they remember about quadratic functions in small groups on poster paper. Each group will then share what they have written down. Using this as a springboard, the definition of a polynomial function will be given. Definition of a polynomial function Let n be a nonnegative integer and let a n,a n 1,,a 2,a 1,a 0 be real numbers with a n 0. The function f ( x) = a n x n + a n 1 x n 1 + + a 2 x 2 + a 1 x + a 0 is called a polynomial function of x with degree n. This should lend to a discussion of how polynomials are classified by their degree or largest exponent. These degreed create certain extended families of polynomials which the students will see when answering the question How many polynomials of degree 5 are there? After they see there are an infinite number of these they should extend this thought to include the idea that there are an infinite number of polynomials of every degree. The polynomial function f ( x) = a, a 0 has degree 0 and is called a constant function. Students should remember that this type of function produces a graph that is a horizontal line intersecting the y-axis at a. Point out that it is a degree zero because the associated x, even though you do not see it, has an exponent of 0. The polynomial function f ( x) = mx + b, m 0 has degree 1 and is called a linear function. This produces a graph that is a line with slope m and y-intercept (0, b). In this lesson the focus is on second degree polynomials, which are called quadratic functions. Definition of a quadratic function Let a, b, and c be real numbers with a 0. The function ( ) = ax 2 + bx + c f x is called a quadratic function. There are numerous examples of quadratics being used to model real life situations, the most common being the quadratic function used to measure the height of an object being fired or dropped from a given height. This function is given by h t where h t ( ) = 16t 2 + v 0 t + h 0 ( ) represents the height of the object at a given time t, v 0 represents the initial velocity or speed of the object, and h 0 represents the initial height of the object.
Pre Calculus 2 When we graph quadratic equations we get a U-shaped curve called a parabola. All parabolas are symmetrical about a vertical line called the axis of symmetry. This axis splits the parabola into two congruent line segments that are mirror reflections of each other. The point where the axis intersects the parabola is called the vertex of the parabola. If the leading coefficient a is positive, then the graph of f x ( ) = ax 2 + bx + c opens up and is a minimum, or lowest point on ( ) = ax 2 + bx + c opens the graph. If the leading coefficient a is negative, then the graph of f x down and is a maximum, or highest point on the graph. When sketching graphs of quadratics it is very helpful to use the graph of the parent function f x ( ) = x 2. You can then use the transformations learned in the last unit to sketch a fairly accurate graph. Remember that y = f x h y = f x ( ) produces a horizontal shift of h units, ( ) + k produces a vertical shift of k units, ( ) produces a vertical stretch with a factor of a, ( ) produces a reflection in the x-axis, and ( ) produces a reflection in the y-axis. y = a f x y = f x y = f x In the previously formed groups, students will be assigned 1 problem from the first 8 problems on page 143 and 1 problem from the numbers 9-12 on the same page. Students will solve problems and then share answers. Students are now ready to be introduced to the standard form of a quadratic equation. Standard form of a quadratic function The quadratic function f x ( ) = a( x h) 2 + k, a 0 is said to be in the standard form. The graph of f is a parabola whose axis is the vertical line x = h and whose vertex is the point (h, k). If a > 0, the parabola opens upward, and if a < 0, the parabola opens downward. This form can come in extremely handy when sketching quadratics because it easily identifies the vertex of the parabola. To demonstrate the usefulness of this form, do problem 31 on page 143 as an example. The process of completing the square will need to be reviewed in this problem.
Pre Calculus 3 ( ) = 2x 2 16x + 31 ( ) = 2( x 2 8x) + 31 ( ) = 2(x 2 8x + 16 16) + 31 ( ) = 2 x 2 8x + 16 ( ) + 31 f x f x f x f x ( ) + 2 16 f ( x) = 2( x 4) 2 1 It is now easy to see that the graph of f is a parabola that opens up with vertex (4, 1). To find the x-intercepts it is important the students understand what is happening at these points. The graph is crossing the x-axis which means y is equal to 0. Therefore, to solve for these intercepts we need to set f x ( ) = 0 and solve for x. The quadratic formula comes in handy when the quadratic is not easily factored. 0 = 2x 2 16x + 31 x = b ± x = b2 4ac 2a 16 ± 256 248 4 ( ) ± 16 2( 2) = 16 = 4 ± 2 2 ( )2 4 2 ( )( 31) Students should note that because of the symmetry of the parabola, x-intercepts are equidistant from the axis of symmetry. This can sometimes be helpful in solving certain problems. If the students are confused by the process of completing the square do a few more as examples and then give them one to try on their own. To find the equation of a parabola, all we need are two points: the vertex and any other point. Once we have these, we insert the vertex values into the standard form and then plug in the values of the other point for x and y. This allows us to solve for a and thus complete our standard form equation. Demonstrate this by doing 37 on page 144 as an example. I first notice that the vertex is at ( 1, 4). I then choose either one of the other points, in this example I will use (1, 0). I then plug in the values for h, k, x, and y to solve for a and get my answer. f ( x) = a( x h) 2 + k 0 = a( 1 ( 1) ) 2 + 4 0 = a( 2) 2 + 4 4 = 4a a = 1 This leaves me with my answer when I plug in 1 for a in the standard form.
Pre Calculus 4 ( ) 2 + 4 f ( x) = 1 x ( 1) f ( x) = ( x + 1) 2 + 4 As much of beginning Calculus deals with finding maxima and minima, many applications involve finding maximum and minimum values of quadratic functions. By writing the quadratic function f x ( ) = ax 2 + bx + c in standard form we can derive an easy way to find the equation for the axis of symmetry and hence the vertex after some simple math. f x ( ) = ax 2 + bx + c ( ) = a x 2 + b a x f x f x + c ( ) = a x 2 + b a x + b2 ( ) = a x 2 + b a x + b2 f x f x ( ) = a x + b 2a 2 4a b2 2 4a 2 + c 4a 2 + c b2 4a + c b2 4a From this we can see that the x-coordinate of the vertex and the equation for the line of symmetry is x = b. This leads to the following generalizations: 2a If a > 0, f has a minimum that occurs at x = b 2a, and If a < 0, f has a maximum that occurs at x = b 2a. To demonstrate how to use quadratics in application problems, do 64, 66, and 76 as examples. 64. First, write an equation with the given information. x + 3y = 42 I can then rewrite y in terms of x. y = 42 x 3 I can then multiply this expression by x and find the maximum value of the numbers by finding the value of x at the vertex/axis of symmetry. x 42 x 3 = 1 3 x2 + 14x I then use the above formula to solve for x.
Pre Calculus 5 x = b 2a = 14 2 1 3 14 3 x = = 21 2 To find y I just have to plug this value for x into the above equation for y which yields 7. 66. (a) P = 2x + 2y 36 = 2x + 2y 36 2x y = = 18 x 2 This means the area will be the product of x and this expression for y. A x ( ) = x( 18 x) The domain for the equation is thus all real numbers, but we know there cannot be a negative or zero area, so the domain for our problem is x 0 < x < 18 { }. (b) (c) It appears that the vertex occurs at x = 9, which would give me a length of 9 as well. This is verified algebraically.
Pre Calculus 6 76. (a) (b) Graphically it appears to be 70. Now we can verify algebraically by plugging in 10 for y. 10 = 0.002s 2 + 0.005s 0.029 0 = 0.002s 2 + 0.005s 10.029 0 = 2s 2 + 5s 10029 5 ± 25 4 2 s = 2 2 ( )( 10029) ( ) 5 ± 80257 s = = 278.296664 = 69.574166 4 4 Very close to my estimation. Concept check: If the graph of a parabola has one x-intercept at (4, 0) and an axis of symmetry at x = 1, where is the other x-intercept located? Where is the vertex located? Homework: pp. 143-146; 15, 17, 21, 27, 29, 35, 41, 43, 47, 53, 57, 61, 65, 69, 73, 77, 79, 81.
Pre Calculus 7 Practice Problems 2.1A: QUADRATICS Name: Determine the vertex of the following quadratics and then tell if it is a minimum or maximum. 1. f ( x) = x 2 14x 44 2. f ( x) = 2( x + 1) 2 + 1 ( ) 3. f ( x) = 9x 2 18x + 77 4. f ( x) = ( x 3) 2 2 Solve the following application problems. To analyze the next problem, we will be using a formula for a freely falling body in which we can ignore any effects of air resistance. s( t) represents the projectile's instantaneous height at any time t v 0 represents initial velocity s 0 represents the initial height from which the projectile is released t represents time in seconds after the projectile is released 5. Some fireworks are fired vertically into the air from the ground at an initial velocity of 80 feet per second. Find the highest point reached by the projectile just as it explodes. 6. The owner of a ranch decides to enclose a rectangular region with 140 feet of fencing. To help the fencing cover more land, he plans to use one side of his barn as part of the enclosed region. What is the maximum area the rancher can enclose? 7. The number of bacteria in a refrigerated food is given by N ( T ) = 20T 2 20T + 120, for 2 T 14 and where T is the temperature of the food in Celsius. At what temperature will the number of bacteria be minimal?
Pre Calculus 8 Practice Problems 2.1A: QUADRATICS (ANSWERS) Name: Determine the vertex of the following quadratics and then tell if it is a minimum or maximum. 1. f ( x) = x 2 14x 44 2. f ( x) = 2( x + 1) 2 + 1 Vertex: (7, 93), min Vertex: (-1, 1), max ( ) 3. f ( x) = 9x 2 18x + 77 4. f ( x) = ( x 3) 2 2 Vertex: (1, 68), min Vertex: (3, 2), min Solve the following application problems. To analyze the next problem, we will be using a formula for a freely falling body in which we can ignore any effects of air resistance. s( t) represents the projectile's instantaneous height at any time t v 0 represents initial velocity s 0 represents the initial height from which the projectile is released t represents time in seconds after the projectile is released 5. Some fireworks are fired vertically into the air from the ground at an initial velocity of 80 feet per second. Find the highest point reached by the projectile just as it explodes. v 0 = 80, s 0 = 0 s t ( ) = 16t 2 + 80t t = b 2a = 80 2 16 ( ) = 80 32 = 2.5 s 2.5 ( ) = 16 2.5 ( )2 + 80 2.5 ( ) = 102.5 ft 6. The owner of a ranch decides to enclose a rectangular region with 140 feet of fencing. To help the fencing cover more land, he plans to use one side of his barn as part of the enclosed region. What is the maximum area the rancher can enclose? P = 2l + w, w = P 2l A = lw = l P 2l l = b 2a = 140 2 2 ( ) = 140 4 = 35 A = lw = 35 140 2 35 ( ) = l(140 2l) = 2l 2 + 140l ( ( ) ) = 35( 70) = 2450 ft2 7. The number of bacteria in a refrigerated food is given by N ( T ) = 20T 2 20T + 120, for 2 T 14 and where T is the temperature of the food in Celsius. At what temperature will the number of bacteria be minimal? T = b 2a = 20 T 1 2 20 2 = 20 1 2 ( ) = 1 2 2 20 1 2 + 120 = 115 bacteria
Pre Calculus 9 Polynomial and Rational Functions: Quiz 1 Name: Objective 1: Describe and define polynomial functions. For the following 2 questions use the polynomial f x ( ) = 5x 5 + 2x 2 x + 13. 1. What is the degree of this polynomial? 2. What is the value of a 0, the constant term? 3. How many degree 4 polynomials are possible? How do you know this? Objective 2: Quadratic Functions 4. Which of the following equations best describes the following graph? a. f x c. f x ( ) = 4(x + 2) 2 + 3 b. f ( x) = 4(x + 2) 2 3 ( ) = 4(x 2) 2 + 3 d. f ( x) = 4(x + 2) 2 + 3 5. Consider the quadratic function f ( x) = 3x 2 + 12x 5. a. What are the x and y intercepts of the graph of this equation? b. What is the standard (vertex) form of this equation? c. What are the coordinates for the vertex of the graph of this equation? d. Is the vertex a maximum or a minimum point? 6. The value of Jon s stock portfolio is given by the function v( t) = 50 + 77t + 3t 2 where v is the value of the portfolio in hundreds of dollars and t is the time in months. How much money did Jon start with? What is the minimum value of Jon s portfolio? 7. The height h in feet of a projectile launched vertically upward from the top of a 96-foot tall tower when time t = 0 is given by h t ( ) = 96 + 80t 16t 2. How long will it take the projectile to strike the ground? What is the maximum height that the projectile reaches? 8. Two rectangular lots are to be made from 400 ft of fencing as seen below. Determine the dimensions that will produce the maximum area. Extra Credit: Show that among all rectangles of fixed perimeter p the one with the largest area is a square.
Pre Calculus 10 Polynomial and Rational Functions: Quiz 1 Name: *****ANSWERS***** Objective 1: Describe and define polynomial functions. For the following 2 questions use the polynomial f x ( ) = 5x 5 + 2x 2 x + 13. 1. What is the degree of this polynomial? Degree 5. 2. What is the value of a 0, the constant term? a 0 = 13. 3. How many degree 4 polynomials are possible? How do you know this? There are an infinite number. You know this because the coefficients can be any real number, and there an infinite number of real numbers. Objective 2: Quadratic Functions 4. Which of the following equations best describes the following graph? a. f x c. f x ( ) = 4(x + 2) 2 + 3 b. f ( x) = 4(x + 2) 2 3 ( ) = 4(x 2) 2 + 3 d. f ( x) = 4(x + 2) 2 + 3 5. Consider the quadratic function f ( x) = 3x 2 + 12x 5. a. What are the x and y intercepts of the graph of this equation? x = 2 ± 21 3 = 0.5 and 3.5, y = 5. b. What is the standard (vertex) form of this equation? f ( x) = 3( x 2) 2 + 7. c. What are the coordinates for the vertex of the graph of this equation? ( 2, 7 ). d. Is the vertex a maximum or a minimum point? Maximum Point. 6. The value of Jon s stock portfolio is given by the function v( t) = 50 + 77t + 3t 2 where v is the value of the portfolio in hundreds of dollars and t is the time in months. How much money did Jon start with? What is the minimum value of Jon s portfolio? Jon started with $50. The minimum value is $50 because t 0. 7. The height h in feet of a projectile launched vertically upward from the top of a 96-foot tall tower when time t = 0 is given by h t ( ) = 96 + 80t 16t 2. How long will it take the projectile to strike the ground? What is the maximum height that the projectile reaches? It will take 6 seconds to reach the ground. The maximum height is 196 feet. 8. Two rectangular lots are to be made from 400 ft of fencing as seen below. Determine the dimensions that will produce the maximum area. Each rectangle will have dimensions 66.7 ft X 50 ft. Extra Credit: Show that among all rectangles of fixed perimeter p the one with the largest area is a square.
Pre Calculus 11 Day 2: Sketching the graph of a polynomial function. Students will be able to describe and sketch the general shape of the graph of a polynomial function. Guiding Questions: When we talk about the end behavior of a graph, what does this mean with reference to the input and output of the polynomial function? What do turning points represent with regards to the output of a function? Why do even roots merely touch the x-axis while odd roots go through the x-axis? One important aspect of intelligent discussion of polynomials is the ability to describe what the general shape of the graph looks like. Not only does this allow one to understand the overall idea of what the polynomial is doing, but it also can shed great light on the relationship between the input and the output of the polynomial. When attempting to describe the shape of a polynomial s graph, there are three important aspects to look at: 1. The end behavior of the graph. 2. The possible number of turning points of the graph of a polynomial. 3. The number and type of roots or x-intercepts. End Behavior The end behavior of the graph of a polynomial describes what the output of a polynomial is as the input tends towards negative infinity and positive infinity. When dealing with polynomials this is determined by the degree of the polynomial and the leading coefficient. Even degree polynomials, by their nature, will always produce outputs that tend to get bigger as the inputs get both big and small when the leading coefficient is positive. When the leading coefficient is negative, or we take the opposite of the polynomial, the trend reverses and the output gets smaller as the input increases and decreases. Odd degree polynomials are different by their nature. When the leading coefficient is positive, they make outputs larger as the inputs increase, and make outputs smaller as the inputs decrease. The above can be summarized by the following table: End Behavior of Polynomial Graphs Even degree polynomial Odd degree polynomial Positive leading coefficient Left side goes up Right side goes up Left side goes down Right side goes up Negative leading coefficient Left side goes down Right side goes down Left side goes up Right side goes down Turning Points When we look at the shape of polynomial graphs, the hills and valleys that appear are very telling as to what the polynomial is doing to the input. These hills and valleys are called turning points, because they are where the graph changes directions and begins a new path. You can understand this by imagining you are driving a car along the path of the
Pre Calculus 12 graph. Turning points would be those places where you would turn the wheel of the car or change the general direction of your path. There are two types of turning points, one where the graph switches directions with respect to the y-axis, and one where the general trend or curve of the graph switches. While predicting the exact number and location of turning points is difficult, it is rather easy to predict the maximum number of turning points the graph of a polynomial might have. The maximum number of turning points is given by the degree of the polynomial minus one. Maximum Number of Turning Points for the Graph of a Polynomial The maximum number of turning points for the graph of a polynomial, f x ( ) = a n x n + a n 1 x n 1 + + a 2 x 2 + a 1 x + a 0, is given by the equation where n is the degree of the polynomial. Maximum number of turning points = n 1 The Roots of a Polynomial The roots of a polynomial are simply the x-intercepts of the graph, where the output is equal to zero (0). These intercepts can have one of two forms: going through the x-axis or touching the x-axis. Going through the x-axis For the intercept to go through the x-axis their needs to be an odd root. This means that there are an odd number of factors that are the same when the polynomial is equal to zero. For the intercept to touch the x-axis their needs to be an even root. Similarly, this means there are an even number of factors that are the same when the polynomial is equal to zero. The way we determine the number of factors that are the same is to look at the exponent associated with a factor when the polynomial is equal to 0. If we look at the following example f ( x) = x 2 ( x + 4) 5 When we set this polynomial equal to 0, there are two roots at x = 0 because the exponent associated with the factor x is a 2. Because this is an even number, the graph will touch the x-axis at this point. There are also 5 roots at x + 4 = 0 because the exponent associated with this factor is a 5. Because this is an odd number, the graph will go through the x-axis at this point. To practice this we will work on some example problems (Worksheet 2.2A).
Pre Calculus 13 Worksheet 2.2A: Shape of Polynomial Graphs Name:
Pre Calculus 14
Pre Calculus 15 Worksheet 2.2A: Shape of Polynomial Graphs ******ANSWERS****** Name:
Pre Calculus 16
Pre Calculus 17 Day 3: Real Zeros of a Polynomial Function Students will be able to divide a polynomial using long division. Students will be able to divide a polynomial using synthetic division. Students will be able to find the real zeros of a polynomial using the Rational Zero Test. Guiding Questions: What does the remainder of polynomial division tell you? Why does the Rational Zero Test only give you possible real zeros? Long Division of Polynomials When dividing anything, we are essentially testing whether a certain value is a factor of a value. If it is a factor then the remainder of the division yields a zero, if it is not a factor then the remainder will be some lesser value. This leads to the definition of division by the Division Algorithm. The Division Algorithm If a and b are values such that a > b and b 0, there exist unique values q and r such that a = bq + r where a is the dividend, b is the divisor, q is the quotient, r the remainder, and r < b. If r = 0, then b divides evenly into a and is a factor of a. When dividing polynomials, we follow the same rules as when dividing numbers. As we divide, we look at the first term of the dividend and divisor, and figure out which term we can multiply the first term of the divisor by to match the first term of the dividend. We then multiply the entire divisor by this term and then place the result under the dividend, subtracting like terms. We then repeat the process until we run out of first terms available. The leftover part at this point is the remainder, and can be added to the quotient as a fraction over the divisor. Example: Divide f ( x) = 6x 3 + 16x 4 by 2x 1. 3x 2 + 3x + 11+ 18 2x 2 2x 2 6x 3 + 0x 2 + 16x 4 ( ) 6x 3 6x 2 6x 2 + 16x ( ) 6x 2 6x 22x 4 ( ) 22x 22 18 Notice that we had to insert an x squared term because it did not exist.
Pre Calculus 18 Synthetic Division Long division can be very tedious, so there is a short-cut when the divisor is in the form x k. When doing synthetic division we use the coefficients of the dividend and the value of k in the divisor. We then multiply the leading coefficient of the dividend by the value of k and then subtract this from the coefficient of the second term of the dividend. We do this until completion, the last result from subtraction being our remainder. Example: Divide 3x 3 2x 2 + 3x 4 by x 3 using synthetic division.
Pre Calculus 19 Both of these division processes can be practiced to make certain that it is understood (Worksheet 3.1A). The Rational Zero Test The Rational Zero Test is a simple way to find possible rational roots, or rational zeros of a polynomial with integer coefficients. It accomplishes this by generating a list of possible ratios, or fractions, of factors of the constant term to the factors of the leading coefficient. The Rational Zero Test If the polynomial f ( x) = a n x n + a n 1 x n 1 + + a 2 x 2 + a 1 x + a 0 has integer coefficients, every rational zero of f has the form Rational Zero = p q where p and q have no common factors other than 1, p is a factor of the constant term, and q is a factor of the leading coefficient. To use the Rational Zero Test you begin by listing all the possible rational numbers whose numerators are factors of the constant term and whose denominators are factors of the leading coefficient. (Note: if the constant term and the leading coefficient have a factor or factors in common other than one, you must factor these out of the polynomial before using this test. These two numbers must be what is called relatively prime.) Possible rational zeros = factors of constant term factors of leading coefficient After you have a list of possible rational zeros (don t forget about negative factors!), you can use the graph of the polynomial on your calculator to narrow the list down to the most likely candidates. Once you have this final list, you can use synthetic division to test if they are indeed factors of the larger polynomial. It is important to note that this list is made up of only possible rational zeros. It is not for certain that this list will contain all, if any, of the real zeros of the polynomial. This is why it is important to use your calculator to test which ones might work, and then verify this by using synthetic division. We will look at example 8 and 10 on pages 167 and 169 respectively to begin. Then you will have homework to look at (Worksheet 3-3B).
Pre Calculus 20 Worksheet 3-3A: Division of Polynomials Name: Divide each polynomial by the given divisor and determine if it is a zero.
Pre Calculus 21 Worksheet 3-3B: Rational Zero Test Name:
Pre Calculus 22 Worksheet 3-3B: Rational Zero Test Name: ******ANSWERS******
Pre Calculus 23 Polynomial and Rational Functions: Quiz 1 Name: Objective 1: Describe and Draw Graph of Polynomials Describe the end behavior of each function. 1. f ( x) = 3x 5 4x 2 + 6 2. g( x) = 8x 10 4x 5 + 2x 1 Find the maximum number of turning points for the graph of each polynomial. 3. f ( x) = 7x 7 3x 3 + x + 4 4. f ( x) = 2x 5 + 3x 2 5 Sketch the graph of the following polynomial. 5. f ( x) = 2x 4 + 3x 3 + 7x 4 Objective 2: Determine Rational Zeros Divide the following polynomials and determine if the divisor is a zero. 6. ( x 4 + 2x 3 2x 2 + 8) ( x + 2) 7. ( 3x 3 x 2 + 4x + 6) ( 3x 12) Determine the possible rational zeros of the polynomial and then find the zeros. 8. f ( x) = 4x 4 + 17x 2 4
Pre Calculus 24 Polynomial and Rational Functions: Quiz 1 Name: Objective 1: Describe and Draw Graph of Polynomials Describe the end behavior of each function. 1. f ( x) = 3x 5 4x 2 + 6 2. g( x) = 8x 10 4x 5 + 2x 1 The left side is pointing up. The right side is pointing down. The left side is pointing up. The right side is pointing up. Find the maximum number of turning points for the graph of each polynomial. 3. f ( x) = 7x 7 3x 3 + x + 4 4. f ( x) = 2x 5 + 3x 2 5 Max Turning Pts = 6 Max Turning Pts = 4 Sketch the graph of the following polynomial. 5. f ( x) = 2x 4 + 3x 3 + 7x 4 Objective 2: Determine Rational Zeros Divide the following polynomials and determine if the divisor is a zero. 6. ( x 4 + 2x 3 2x 2 + 8) ( x + 2) 7. ( 3x 3 x 2 + 4x + 6) ( 3x 12) x 3 2x + 4, Is a zero. x 2 + 11 198 x + 16 +, Not a zero. 3 3x 12 Determine the possible rational zeros of the polynomial and then find the zeros. 8. f ( x) = 4x 4 + 17x 2 4 PRZ: ± 1 1,± 1 2,± 1 4,± 2 1,± 4 1. Actual zeros: 2, 1 2, 1 2,2.
Pre Calculus 25 Day 4: Complex Numbers Students will be able to add, subtract, multiply, and divide complex numbers. Guiding questions: What must you multiply a complex number by to get rid of the imaginary part? As you raise the imaginary number i to successively higher exponents, what patterns emerge? The Mandelbrot Set is composed of numbers which are bounded by the formula that produces it. What is meant by bounded? Complex Numbers and the Complex Plane Complex numbers are not as difficult as the name might suggest. These numbers are composed of two parts; one real and one imaginary. Definition of a Complex Number The standard form of a complex number is given by a + bi Where a and b are real numbers, and i is the imaginary unit, equal to the square root of negative one. The first part, a, is considered the real part. The second part, bi, is considered the imaginary part. Complex numbers is the parent set to both the real and imaginary numbers. This can be shown by setting a and b equal to zero independently. If a = 0, then we have an imaginary number. While if b = 0, we have a real number. Complex numbers are made by simply substituting real number values in for a and b. For example, if I set a = 2 and b = 3, the resulting complex number C would be 2 + 3i. Complex numbers can be plotted on the complex plane, the perpendicular intersection of the horizontal real number line and the vertical imaginary number line at their zeros. If we plot our above complex number on this plane it would look like this: As you can see, to plot this number we move horizontally along the real number axis 2 units, and then up from this point 3i units parallel to the imaginary number line.
Pre Calculus 26 Operations With Complex Numbers Addition and Subtraction To add or subtract complex numbers we combine like terms. This means that we add or subtract the two real parts, and add or subtract the two imaginary parts. Addition and Subtraction of Complex Numbers If a + bi and c + di are two complex numbers, we add and subtract them as follows. Addition: ( a + bi) + ( c + di) = ( a + c) + ( b + d)i Subtraction: ( a + bi) ( c + di) = ( a c) + ( b d)i Example: Simplify the following ( 2 4i) + i ( 5 + 3i) = 2 4i + i 5 3i Remove the parentheses (don t forget to distribute the negative sign!) = 2 5 4i + i 3i Group like terms. = ( 2 5) + ( 4 + 1 3)i Combine like terms. = 3 6i Write in standard form. Multiplication When we multiply complex numbers we use distribution just as we would with any set of binomials. Multiplying Complex Numbers If a + bi and c + di are two complex numbers, we add and subtract them as follows. ( a + bi) ( c + di) = a( c + di) + bi( c + di) = ac + ( ad)i + ( bc)i + ( bd)i 2 = ac + ( ad)i + ( bc)i + bd = ac bd + ( ad)i + ( bc)i = ( ac bd) + ( ad + bc)i ( )( 1) Distributive property. Distributive property. i 2 = 1 Commutative property. Combine like terms (reverse distributive). Example: Simplify the following 3 2i ( )( 1+ 3i) ( 3 2i) ( 1+ 3i) = 3 + 9i 2i 6i 2 = 3 + 9i 2i 6( 1) = 3 + 6 + 9i 2i = 9 + 7i
Pre Calculus 27 Division When we are dividing complex numbers, we have to get a denominator that has no imaginary part. To do this, we multiply by the complex conjugate of the denominator. Complex conjugates are pairs of complex numbers that have equal real parts and opposite imaginary parts. Complex conjugates take the form of a + bi and a bi. When we multiply these together they will always produce a real number. Proof: ( a + bi) ( a bi) = a 2 abi + abi b 2 i 2 = a 2 b 2 ( 1) = a 2 + b 2 Dividing Complex Numbers To find the quotient of a + bi and c + di where c and d are not both zero, multiply the numerator and denominator by the conjugate of the denominator. ( a + bi) c + di ( ) = ( a + bi) ( c + di) ( c di) ( c di) ( ac + bd) + bc ad = ( )i c 2 + d 2 Example: Simplify the following 3 i 2 + 3i 3 i 2 + 3i = 3 i 2 + 3i ( ) ( ) ( 2 3i) ( 2 3i) 6 9i 2i + 3i2 = 4 9i 2 6 3 9i 2i = 4 + 9 = ( 3 11i ) 13 = 3 13 11 13 i The homework (2-4A) will allow you to practice these as well as investigate a few more interesting aspects and applications of complex numbers, including FRACTALS!
Pre Calculus 28 Homework 2-4A: Complex Numbers Name: Part I: Practice Problems Graph the following complex numbers on the given complex plane. 1. 1 4i 2. 2 + i 3. 3 4. 2i 5. i( 2 i) Find the conjugate of the following. 6. 1+ i 7. 4 3i 8. 2 5i Simplify the following complex expressions. 9. ( 1 5i) + ( 2 + 3i) 10. ( 4 + 2i) ( 3 4i) 11. i + ( 2 3i) ( 5 6i) 12. ( 2 + i) ( 1 4i) 13. ( 2 + 3i) ( 4 + 2i) 14. i( 1+ 5i) ( 1 5i) 15. 2 + 3i 1+ i 16. 4 5i 3 2i 17. 3 + i 7 2i
Pre Calculus 29 Part II: Exploration Investigation: Degrees of i Complete the following table: i 1 = i i 2 = 1 i 3 = i i 4 = 1 i 5 = i 6 = i 7 = i 8 = i 9 = i 10 = i 11 = i 12 = Question What pattern do you see? Write a brief description of how you would find i raised to any positive integer power. Extension What kind of pattern would you expect to see when raising i to negative integer powers? Write a brief description of what you find including any work you do.
Pre Calculus 30 Activity: The Mandelbrot Set Fractals are at the core of many modern day technological advancements. Fractals are endlessly repeating patterns that have an infinite amount of similarity both big and small. The most famous fractal pattern is called the Mandelbrot Set, named after the mathematician Benoit Mandelbrot. The Mandelbrot Set is a recursive function defined as follows: 2 ( ) = z n 1 f z n + z 0 where z 0 is a complex number that we choose. For some values of z 0 this sequence is bounded, which means that all elements in this sequence are less than some fixed number N. We call these sequences prisoners, because they cannot escape the boundary of that fixed value. Other values, however, are unbounded, which means that the elements in these sequences keep growing and become infinitely large. If the sequence is bounded, then the complex number we initially chose, z 0, is in the Mandelbrot Set; if the sequence is unbounded, the complex number z 0 is not in the Mandelbrot Set. For example, let s say we chose a value for z 0 = i. The first few values in this sequence are: z 0 = i z 1 = i 2 + i = 1+ i ( ) 2 + i = i ( ) 2 + i = 1+ i ( ) 2 + i = i ( ) 2 + i = 1+ i ( ) 2 + i = i z 2 = 1+ i z 3 = i z 4 = 1+ i z 5 = i z 6 = 1+ i You can see this sequence gets stuck, so to speak. This makes it bounded, because it is a prisoner of these fixed values. Therefore we could plot this initial complex number, z 0 = i, on the complex plane as part of the Mandelbrot Set. The complete set of points in the Mandelbrot Set looks like this: Challenge Extension Find two complex numbers that are part of the Mandelbrot Set, and two that are not part of the Mandelbrot Set. Show your work and justification for each. Find two applications of fractals in modern day technology. Research what role fractals play in these applications and present your findings in a short 1 page paper.
Pre Calculus 31 Homework 2-4A: Complex Numbers Name: ******ANSWERS****** Part I: Practice Problems Graph the following complex numbers on the given complex plane. 1. 1 4i 2. 2 + i 3. 3 4. 2i 5. i( 2 i) Find the conjugate of the following. 6. 1+ i 7. 4 3i 8. 2 5i 1 i 4 + 3i 2 + 5i Simplify the following complex expressions. 9. ( 1 5i) + ( 2 + 3i) 10. ( 4 + 2i) ( 3 4i) 11. i + ( 2 3i) ( 5 6i) 3 2i 1+ 6i 7 + 4i 12. ( 2 + i) ( 1 4i) 13. ( 2 + 3i) ( 4 + 2i) 14. i( 1+ 5i) ( 1 5i) 6 7i 14 + 8i 26i 15. 2 + 3i 1+ i 16. 4 5i 3 2i 17. 3 + i 7 2i 5 2 + 1 2 i 22 13 7 13 i 19 53 + 13 53 i
Pre Calculus 32 Day 5: The Fundamental Theorem of Algebra Students will be able to find the zeros of a polynomial function. Guided questions: How many zeros does a polynomial have? Why do imaginary roots come in pairs? What do we call these pairs? The Fundamental Theorem of Algebra is so fundamental that it is extremely easy to understand. It says that the total number of zeros that a polynomial has in the complex number system is equal to the degree of the polynomial. So, if the degree of a polynomial is 2, it has 2 roots. If it is a 5 th degree polynomial, it has 5 roots. The secret to finding the zeros of a polynomial is factoring, namely being able to factor using complex numbers. Specifically, the special factor pattern found with the difference of squares is particularly handy, as it allows one to easily find imaginary roots. Let s look again at the difference of squares pattern. a 2 b 2 = a + b ( )(a b) Looking at this, what does it remind you of from the section we just studied about complex numbers? It is suspiciously similar to the form of complex conjugates. In fact, this is it; complex conjugates are a special case of the difference of squares. This leads to an important generalization about the zeros of a polynomial, that all complex zeros occur in conjugate pairs. Complex Zeros Occur In Conjugate Pairs Let f ( x) be a polynomial function that has real coefficients. If a + bi, where b 0, is a zero of the function, the conjugate a bi is also a zero of the function. This leads to the last of the important concepts, irreducibility. If a polynomial can be factored so that all zeros are real, then it is reducible over the rationals. If it can be factored, but at least one of the zeros is real but not rational; we say this polynomial is irreducible over the rationals, but reducible over the reals. Finally, if it can be factored, but has at least one zero that contains an imaginary number; we say this polynomial is irreducible over the reals. The point of factoring a polynomial is to reduce a complicated higher degree algebraic expression to a string of linear factors which will be easy to find the zeros of. It will be assumed that you can already perform factoring under the set of real numbers, so we will look at examples of factoring under the complex number system when imaginary numbers are involved. Example 1: Factor the polynomial x 2 + 5. Solution: To factor this we need to rewrite it as the difference of two squares. x 2 5 ( ) We now can use the special factor pattern to simplify it. ( x + 5 )( x 5 ) We now can pull out the imaginary numbers to get our answer. ( x + 5i) ( x 5i) QED You will be able to practice this on the worksheet given out today (Worksheet 2-5A).
Pre Calculus 33 What if we are given the zeros? How do we write the polynomial given this information? Example 2: Find a third degree polynomial function, with real coefficients, that has 2 and 4i as zeros. Solution: The first thing we do is find the linear factors that match our zeros. To do this we subtract the zeros from x. This gives us the following. ( x 2), ( x 4i) But wait! Remember that imaginary roots come in complex conjugates, so we will have to include the conjugate of x 4i ( ), ( x + 4i) So, our completed list of linear factors is as follows, ( x 2), ( x 4i), x + 4i ( ) Since these are the factors of our polynomial, we will need to multiply them together to have the original form of the polynomial reveal itself. f x ( ) = a( x 2) ( x 4i) ( x + 4i) Notice the a in front of the linear factors. To make our problem simple, we will always let a = 1, to obtain f x ( ) = ( x 2) ( x 4i) ( x + 4i) = ( x 2) x 2 + 16 ( ) = x 3 2x 2 + 16x 32 QED When we are asked to find the zeros of a polynomial on our own, we have to rely on the tools we have begun to develop this unit; using the calculator, the rational zero test, and synthetic division. Example 3: Find all zeros of f ( x) = x 5 + x 3 + 2x 2 12x + 8. Solution: Using the rational zero test we find that the possible rational zeros are PRZ's = ±1,±2,±4,±8 Looking at the graph of the polynomial on the calculator we can see that 2 and 1 are good guesses (with 1 being a probable repeated zero). Using synthetic division, we can determine that 2 is indeed a zero and 1 is a zero twice, so we can rewrite the polynomial function as By factoring x 2 + 4 as x 2 4 f ( x) = ( x 1) ( x 1) ( x + 2) ( x 2 + 4) ( )( x + 4 ) = x 2i ( ) = x 4 we get ( )( x + 2i) f ( x) = ( x 1) ( x 1) ( x + 2) ( x 2i) ( x + 2i) which gives the following five zeros of f. 1, 1, 2, 2i, and 2i Notice that on the graph of the previous example s polynomial that only the real zeros are visible, or discernible as actual x-intercepts.
Pre Calculus 34 Worksheet 2-5A: Fundamental Theorem of Algebra Name: Determine the number of zeros that each polynomial has. 1. f ( x) = 3x 6 + 5x 3 2x 2. f ( x) = x 3 x + 5 3. f ( x) = 4x 4 + 2x 3 6x + 1 Find all the zeros of the function. 4. f ( x) = x( x 6) 2 5. f ( x) = x 2 ( x + 3) ( x 2 1) 6. f ( x) = ( x + 6) ( x + i) ( x i) Factor the following polynomials using the difference of squares pattern and find the zeros. 7. f ( x) = x 2 + 16 8. f ( x) = x 2 + 7 9. f ( x) = x 4 81 Find a polynomial function with integer coefficients that has the given zeros. 10. 1, 5i, 5i 11.4, 3i 12.2, 4 + i, 4 i Find the zeros of the following polynomials. 13. f ( x) = x 3 3x 2 + 4x 2 14. f ( x) = x 4 6x 2 7 Use the given zero to find all the zeros of the function. 15. Function: f ( x) = 2x 3 + 3x 2 + 50x + 75 Zero: 5i 16. Function: f ( x) = x 3 7x 2 x + 87 Zero: 5+2i 17. The demand equation for a cell phone is p = 140 0.0001x where p is the unit price (on dollars) of the phone and x is the number of units produced and sold. The cost equation for the phone is C = 80x + 150,000 where C is the total cost (in dollars) and x is the number of units produced. The total profit obtained by producing and selling x units is P = R C = xp C. If you were consulted by a company with this information, and were asked to determine a price p that would yield a profit of 9 million dollars, would that be possible? Explain you response to this company s request. 18. Is it possible for a third degree polynomial function with integer coefficients to have no real zeros? 19. For the function f ( x) = x 4 4x 2 + k, for what values of k would f have (a) four real zeros, (b) two real zeros each of multiplicity 2, (c) two real zeros and two complex zeros, and (d) four complex zeros.
Pre Calculus 35 Day 6: Rational Functions and Asymptotes Students will be able to find the domains of rational functions, and find horizontal and vertical asymptotes of rational functions. Guiding questions: How is the domain related to the asymptotes of a rational function? How do the degrees of the polynomials in a rational function help determine asymptotes of that rational function? A rational function is a ratio composed of two polynomials, much like a rational number is a ratio composed of two integers. It can be written in the form ( ) = N ( x) D( x) f x where N ( x) and D( x) are polynomials, D( x) 0, and N ( x) and D( x) have no common factors. Notice that the polynomial in the numerator is labeled N and the polynomial in the denominator is labeled D. Also, when we say these two polynomials have no common factors, we simply mean that f is in simplest form. When we talk about the domain of a rational function, in general we are referring to all input values for x that do not make the denominator equal to zero. Therefore, when we are asked to find the domain of a rational function, we set the denominator equal to zero and solve for x to determine the domain. While domain may speak to the relationship between input and output algebraically, it has interesting consequences graphically as well. We are going to investigate this relationship (Worksheet 2-6A) and then summarize the results. The goal is to define what the terms horizontal asymptote and vertical asymptote mean visually and algebraically. To summarize our results, Asymptotes of a Rational Function Let f be the rational function ( ) = N ( x) D( x) = a n xn + a n 1 xn 1 + + a 1 x + a 0 b m x m + b m 1 x m 1 + + b 1 x + b 0 ( ) and D( x) have no common factors. f x where N x 1. The graph of f has vertical asymptotes at the zeros of D x 2. The graph of f has at most one horizontal asymptote determined by comparing the degree of N x ( ) and D( x). ( ). a. If n < m, the line y = 0 (the x-axis) is a horizontal asymptote. b. If n = m, the line y = a n is a horizontal asymptote. bm c. If n > m, the graph of f has no horizontal asymptote.
Pre Calculus 36 Now we must put our knowledge into action and try to graph rational functions. To do this we will first work without calculators and investigate the best way to do this by hand. In small groups students will investigate how to graph three rational functions, devising a method and then testing this method for its effectiveness and ease of use (Investigation 2-6B). Exit Slip: It is important to understand what these results mean. Vertical asymptotes are easy because we know you cannot divide by zero. But, what about horizontal asymptotes? See if you can come up with a reason why 2 a, b, and c are always true. What makes these the case?
Pre Calculus 37 Worksheet 2-6A: Domain and Asymptotes Name: A rational function is a ratio composed of two polynomials, much like a rational number is a ratio composed of two integers. It can be written in the form f ( x) = N ( x) D( x) where N ( x) and D( x) are polynomials, D( x) 0, and N ( x) and D x ( ) have no common factors. Notice that the polynomial in the numerator is labeled N and the polynomial in the denominator is labeled D. Also, when we say these two polynomials have no common factors, we simply mean that f is in simplest form. When we talk about the domain of a rational function, in general we are referring to all input values for x that do not make the denominator equal to zero. Therefore, when we are asked to find the domain of a rational function, we set the denominator equal to zero and solve for x to determine the domain. Consider the following rational functions. A) f ( x) = 3x2 2x 1 B) f ( x) = 7 x 2 + 1 C) f ( x) = 5x 1 x + 3 D) f ( x) = 4x4 + 2x 2 1 x 4 16 What is the domain for each function? A) B) C) D) Sketch the graph for each function. Draw vertical and horizontal dashed lines where you see an apparent break in the continuity of the graph. A) B) C) D) What is the relationship between the domain and the vertical breaks in the graph? For each graph, look at the breaks in the graph that are horizontal in nature. What part of the rational function accounts for this? (Hint: Look at the degree and leading coefficients for each polynomial in the rational functions.) Describe what you find. These breaks in the graph are called asymptotes. If you were in charge of defining them for successive math students, what would your definition be?
Pre Calculus 38 Worksheet 2-6A: Domain and Asymptotes Name: ******ANSWERS****** A rational function is a ratio composed of two polynomials, much like a rational number is a ratio composed of two integers. It can be written in the form f ( x) = N ( x) D( x) where N ( x) and D( x) are polynomials, D( x) 0, and N ( x) and D x ( ) have no common factors. Notice that the polynomial in the numerator is labeled N and the polynomial in the denominator is labeled D. Also, when we say these two polynomials have no common factors, we simply mean that f is in simplest form. When we talk about the domain of a rational function, in general we are referring to all input values for x that do not make the denominator equal to zero. Therefore, when we are asked to find the domain of a rational function, we set the denominator equal to zero and solve for x to determine the domain. Consider the following rational functions. A) f ( x) = 3x2 2x 1 B) f ( x) = 7 x 2 + 1 C) f ( x) = 5x 1 x + 3 D) f ( x) = 4x4 + 2x 2 1 x 4 16 What is the domain for each function? A) x 1 2 B) x ±i C) x 3 D) x ±4, ±4i Sketch the graph for each function. Draw vertical and horizontal dashed lines where you see an apparent break in the continuity of the graph. A) B) C) D) What is the relationship between the domain and the vertical breaks in the graph? The vertical breaks occur at the excluded values of the domain. For each graph, look at the breaks in the graph that are horizontal in nature. What part of the rational function accounts for this? (Hint: Look at the degree and leading coefficients for each polynomial in the rational functions.) Describe what you find. If the degree of the polynomial in the numerator is greater than that in the denominator, there is no horizontal break. There does appear to be a break at y = a n xn b m x m. If the degrees of the polynomials in the numerator and denominator are the same, there is a horizontal break at y = a n bm. If the degree of the polynomial in the numerator is less than that in the denominator, there is a horizontal break at y = 0, the x-axis. These breaks in the graph are called asymptotes. If you were in charge of defining them for successive math students, what would your definition be? Answers will vary.
Pre Calculus 39 Investigation 2-6B: Graphing Rational Functions Name: For each rational function, graph it by hand without a calculator. Try to develop steps, or a method, that you could use to graph any rational function. Function Graph of Function f ( x) = 1 x 3 f ( x) = 4 x 1 ( )( x + 2) f ( x) = 3x 2 x + 4 ( )( x + 1) f ( x) = 2x3 x 4 4
Pre Calculus 40 Homework 2-6C: Rational Functions Name: For questions 1-6, find the domain, vertical asymptote(s), horizontal asymptote(s), and then graph the function. 1. f ( x) = 1 x 2 2. f ( x) = 3 ( x 2) 3 3. f ( x) = 3 + x 3 x 4. f ( x) = 2 5x 2 + 2x 5. f ( x) = 2x3 x 2 1 6. f ( x) = 3x2 + 1 x 2 + x + 9 7. The cost (in millions of dollars) for removing p% of the industrial and municipal pollutants discharged into a river is C = 255 p, 0 p < 100. What is the cost of removing 10% of 100 p the pollutants? 40%? 75%? What would be an appropriate viewing window on your calculator when graphing this function? According to this model, would it be possible to remove 100% of the pollutants? 8. A biology class performs an experiment comparing the quantity of food consumed by a certain kind of moth with the quantity supplied. The model for the experimental data is 1.568x 0.001 y =, x > 0 where x is the quantity (in milligrams) of food supplied and y is 6.360x + 1 the quantity (in milligrams) eaten. At what level of consumption will the moth become satiated? 9. True or False? A rational function can have infinitely many vertical asymptotes. 10. Write a rational function that has no vertical asymptote and a horizontal asymptote at y = 2.
Pre Calculus 41 Homework 2-6C: Rational Functions Name: ******ANSWERS****** For questions 1-6, find the domain, vertical asymptote(s), horizontal asymptote(s), and then graph the function. 1. f ( x) = 1 x 2 2. f ( x) = 3 ( x 2) 3 3. f ( x) = 3 + x 3 x V.A.: x=0, H.A.: y=0 V.A.: x=2, H.A.: y=0 V.A.: x=3, H.A.: y= 1 4. f ( x) = 2 5x 5. f ( x) = 2x3 6. f ( x) = 3x2 + 1 x 2 + x + 9 2 + 2x x 2 1 V.A.: x= 1, H.A.: y= 5/2 V.A.: x=1, 1; H.A.: none V.A.: none; H.A.; y=3 7. The cost (in millions of dollars) for removing p% of the industrial and municipal pollutants discharged into a river is C = 255 p, 0 p < 100. What is the cost of removing 10% of 100 p the pollutants? 40%? 75%? What would be an appropriate viewing window on your calculator when graphing this function? According to this model, would it be possible to remove 100% of the pollutants? $28.33 million; $170 million; $765 million; ; No, function is undefined. 8. A biology class performs an experiment comparing the quantity of food consumed by a certain kind of moth with the quantity supplied. The model for the experimental data is 1.568x 0.001 y =, x > 0 where x is the quantity (in milligrams) of food supplied and y is 6.360x + 1 the quantity (in milligrams) eaten. At what level of consumption will the moth become satiated? About 0.247 milligrams. 9. True or False? A rational function can have infinitely many vertical asymptotes. False. A polynomial cannot have infinite degree. 10. Write a rational function that has no vertical asymptote and a horizontal asymptote at y = 2. Example: f ( x) = 2x2 x 2 + 1.
Pre Calculus 42 Unit 2: Rational Functions, Quiz 2 Name: Use the following rational function to answer the next 4 questions. 1. What is the domain of this function? f ( x) = 5x3 1 2x 3 8x 2. List the vertical asymptotes if there are any. 3. List the horizontal asymptote if there is one. 4. Graph the function. 5. Write a rational function that has a vertical asymptote at x = 2, and a horizontal asymptote at y = 5. 6. A friend asks for your help in graphing a rational function. They already have the asymptotes drawn as shown and, before you even see the function, you know they re wrong. How do you know this?
Pre Calculus 43 Unit 2: Rational Functions, Quiz 2 Name: ******ANSWERS****** Use the following rational function to answer the next 4 questions. 1. What is the domain of this function? x x 2,2 { } 2. List the vertical asymptotes if there are any. x = 2, x = 2 3. List the horizontal asymptote if there is one. f ( x) = 5x3 1 2x 3 8x y = 5 2 4. Graph the function. 5. Write a rational function that has a vertical asymptote at x = 2, and a horizontal asymptote at y = 5. Example: f ( x) = 5x x 2 6. A friend asks for your help in graphing a rational function. They already have the asymptotes drawn as shown and, before you even see the function, you know they re wrong. How do you know this? You know this because they have two horizontal asymptotes which is not possible.