College Algebra with Current Interesting Applications and Facts by Acosta & Karwowski, Kendall Hunt, Textbook Supplement (Revised 8/2013)

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1 College Algebra with Current Interesting Applications and Facts b Acosta & Karwowski, Kendall Hunt, 01 Tetbook Supplement (Revised 8/013) Contents A. Comple Numbers B. Distance Formula, Midpoint Formula, and Circles 9 C. Snthetic Division 3 Topic Epansion from Content Alread in Tetbook D. More on Rational Functions 33 E. Non-Linear Sstems of Equations 7 F. Other Tpes of Equations 1 G. Solving Polnomial and Rational Inequalities 71

2 Section A: Comple Numbers Suppose we wish to solve the equation = 5. Since we know that the square of a real number is either zero or positive, there is no real number that would satisf this equation. To solve this problem, mathematicians created a number sstem that is based upon a new number: the imaginar unit, commonl referred to as "i." Little Facts: It is said that the name "imaginar number" was originall coined b René Descartes in the seventeenth centur as a derogator term, because at that time such numbers were regarded b some as fictitious or useless. Swiss mathematician Leonhard Euler introduced the letter "i" to represent the square roots of negatives in In modern times these numbers have essential, concrete applications in math, phsics, electrical engineering, and man other scientific and related areas. Sources: The Imaginar Unit The imaginar unit, i, is defined with the following properties i = 1 and i = 1 1 = 1 We can use the imaginar unit to rewrite the square root of a negative number. Rewriting the Epression If a > 0, then = i Let us rewrite 5 as a product of a real number and i: 5 = ( 1)(5) = i 5 = 5i We can check the answer b squaring 5i: (5i) = 5 i = (5)( 1) = 5 Eample 1 Rewrite 8 in terms of i, and simplif if possible. 8 = i 8 = i 7 = i 7 Note: It is also acceptable to write an epression like i 7 as 7i, but we must be sure to write the "i" outside the radical smbol. To avoid being read as being under the radical, we generall write the answer with "i" in front of the radical. When mathematicians added a real number to multiples of imaginar units, the set of comple numbers was formed. Comple Number A comple number is one of the form a + bi, where a and b are real numbers. In a comple number, we call a the real part and b is the imaginar part. Two comple numbers a + bi and c + di are equal if and onl if a = c and b = d. An real number, a, can be written as a comple number as a + 0i. In this case, b = 0. If a comple number has b 0, then we call a + bi, an imaginar number (nonreal comple

3 number). On the other hand, if b 0 but a = 0, then a + bi = 0 + bi = bi and we call this a pure imaginar number. Some eamples of comple numbers are: + 5i imaginar number, a 0 b 0 7i pure imaginar number, a = 0 1 real number, b = 0 A comple number written in the form a + bi is said to be in standard form. We have seen that the comple number sstem includes the set of real numbers as well as the set of imaginar numbers. The following diagram displas the relationship amongst comple, real, and imaginar numbers, where a and b are real numbers. Some eamples are included. +5, 3, Comple Numbers a + bi +, i, 7, 1, 0,, 3, π, 9, 8 Real Numbers a + bi, b = 0 1, 0,, 3, π, 9, 8 Imaginar Numbers a + bi, b 0 +5, 3, +,, 7 Operations with Comple Numbers All the real properties of real numbers will hold for the comple numbers. We can add, subtract, multipl, and divide comple numbers using the real number properties and the operational definitions that we will discuss in this section. Addition and Subtraction of Comple Numbers For numbers a + bi and c + di, add or subtract the real parts and add or subtract the imaginar parts. (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) (c + di) = (a c) + (b d)i Eample Perform the indicated operation and simplif. Write our answer in standard form. a. ( + 5i) + (7 3i) b. ( 8i) ( + i) c. (3 5i) + (1 + 3i) (17 i) a. ( + 5i) + (7 3i) = ( + 7) + [5 + ( 3)]i = 9 + i 3

4 b. ( 8i) ( + i) = ( ) + ( 8 1)i = 10 9i c. (3 5i) + (1 + 3i) (17 i) = ( ) + ( )i = 13 or i Multiplication of Comple Numbers For comple numbers a + bi and c + di, follow the same rules for multipling binomials, remembering that i = 1. (a + bi)(c + di) = ac + adi + bci + bdi = ac + adi + bci + bd( 1) = (ac bd) + (ad + bc)i The good news is that we do not have to memorize this definition. We simpl multipl the given comple numbers as with binomials. Note: To simplif epressions containing comple numbers, start b rewriting an square roots with negative radicands as pure imaginar numbers. Eample 3 Perform the indicated operation and simplif. a. ( 5i)( + 3i) b. 1 9 c. (3 + 7i) d. (3 + 5i)(3 5i) a. ( 5i) ( + 3i) = 1 + i 30i 15i = 1 i 15( 1) = 7 i b. 1 9 = i 3i = 1i = 1( 1) = 1! CAUTION Recall from elementar algebra that = onl when both a and b are non-negative, thus c. (3 + 7i) = 3 + (3)(7i) + (7i) Square of a binomial = 9 + i + 9( 1) = 0 + i d. (3 + 5i)(3 5i) = 3 (5i) Product of sum and difference of same two terms = 9 5( 1) = 3 From eample 3(d), notice that when we multiplied the given comple numbers we obtained a real number. This will occur when we multipl a pair of comple numbers of the form (a + bi)(a bi), which we call comple conjugates.

5 Propert of Two Comple Conjugates For real numbers a and b, (a + bi)(a bi) = a + b (a bi)(a + bi) = a + b For z = +, some tetbooks use the epression = + to denote the comple conjugate. We will use the aforementioned propert to divide comple numbers. Division of Two Comple Numbers To obtain the quotient of two comple numbers, multipl the numerator and denominator b the comple conjugate of the denominator and simplif. + +! +! = # $# +!! $ Eample Epress each quotient in standard form a + bi. a. & ' ( ' b. ' a. We need to eliminate the imaginar part from the denominator, so that we can epress the quotient in standard form. Multipling both numerator and denominator b the comple conjugate, we will obtain a real number as the new denominator. & ' (' = ) & ' (' *) &' &' & ' & ' & (() = (' + ((() * = & ' & ' & '+ = (& ' = + b. = ' ) ' *)( ' * = (' = (' = (' (' (' + ((() = 9i or 0 9i Eample 5 Divide and simplif. a. (, ( b. - & (. a. We first rewrite the square roots with negative radicands as pure imaginar numbers, then divide and simplif. (, = ', ( ' = /, = 3 = 1 = 5

6 b. - & (. = - & '. = = - & ' 0 = - & ' = - + ' = + i 5 Powers of i From the definition of the imaginar unit, we alread know that that i = 1. Consider the following cclic pattern: = i 5 = i i = 1 i = i = 1 i = i i = 1 ( 1) = 1 i 3 = i i = ( 1) i = i 7 = i i 3 = 1 ( i) = i = i i = ( 1)( 1) = 1 i 8 = i i = 1 1 = 1, and so on. Observe that the powers of i repeat ever fourth power, and the result is alwas one of onl four possibilities:, 1,, or 1. We can use the following steps to simplif higher powers of i. Simplifing powers of i for 1, n > 1. Divide the eponent n b, and check the remainder, r.. Replace 3 with (replace the original power of i with the remainder R) 3. Use the cclic pattern =, i = 1, i 3 =, i =1 as needed to simplif. Note: Recall that i 0 =1 b the zero eponent propert. Eample Simplif each power of i. a. i 13 b. i 30 c. (, a. 0 = 3, remainder R = 1. So, = i. Confirming the answer: i 13 = i 1 i = ( 0 ) 3 i =1 3 i = b.. 0 = 7, R =. So, i = 1. Confirming the answer: i 30 = ( 0 ) 7 i = (1) 7 i =1 ( 1) = 1 c. (, 0 =, R = 0. So, i0 = 1. Optionall, ou can confirm the answer. Graphing Calculator and Comple Numbers Most graphing calculators have a comple numbers mode and we can use this as an optional tool for checking our work with comple numbers. We can use the M ke to change the calculator from REAL numbers mode to a + bi mode. If we wanted to evaluate 9 using the real numbers mode, the calculator will give us an error message, because this is not a real number. Using the comple number mode, the calculator will generate the correct answer, 3i. See the screens that follow.

7 The following screens show the answers we obtained in Eamples a, and b, 3a and 3b, and a. You can confirm the other eamples. In man calculators we can use `. to tpe i.! CAUTION Remember to switch the calculator mode back to REAL numbers once ou are finished working with comple numbers. In-Class Practice Section (A) 1. Rewrite in terms of i, and simplif if possible: a. 81 b. 13 c. 00. Perform the indicated operation and simplif. Write our answer in standard form. a. (5 + i) + (8 i) b. (10 i) ( + 9i) c. (8 3i)(1 1i) d e. (3 5i) f. (.. (0 g., & ' ( ' h. & (5... i. ( 7 i)( 7 i) j. i( +5i) ( 1 ) 3. Simplif each power of i. a. i 19 b. i 0 c. (-. Confirm our answers to problem b and f with our graphing calculator. 7

8 Eercises Section (A) In eercises 1-30, perform the indicated operation and simplif. Write answers in standard form. 1. ( + 8i) + (9 3i). (3 0i) (7 13i) 3. (1 + i) (10 i) 17i. (19 7i) + ( 0 i) ( + 7i) 5. ( 5i)(3 + i). ( 1 i)( 9 i) 7. ( 8 + i)( 7i) 8. (9 + 7i)( 9 7i) 9. ( + 3i) 10. ( 3 7i) 11. ( 5 + i)( 5 i) 1. ( 9i)( +9i) 13. ( 9i) 1. (10 +) i + 3i. 3 i + i i 1+ i. 1 i 7 8i In eercises 31-3, simplif each power of i i i 33. i i i 3. i Answer Ke Section (A) Eercises i i i 7. i 9. + i

9 Section B: Distance Formula, Midpoint Formula, and Circles Distance Formula Suppose we want to determine the distance between the points 7 (3, 0) and 7 (8, 0) in the coordinate plane. B just looking at the graph, we can tell that these points are 5 units apart P 1 P Points 7 and 7 have -coordinate 0, and we know that the lie on the -ais, a horizontal line. Recall that the distance between two points a and b on the number line is!, =. So, the distance between two points on the -ais is simpl 9 9. Likewise, the distance between two points on the -ais, a vertical line, is : :. In our graph, the distance between 7 and 7 is 9 9 = 8 3 = 5. Observe that we are not measuring the distance from point 7 to point 7, which would impl direction. We are interested in finding the distance between the given points, regardless of direction. Thus, this distance can also be found b 9 9 = 3 8 = 5 = 5. So, how do we calculate the distance between an points on the coordinate plane, which we know will not alwas reflect horizontal or vertical lines? We can use the Pthagorean Theorem to develop a formula for the distance between an two points (9, : ) and (9, : ) in the coordinate plane. Let us construct a right triangle with a horizontal leg of length 9 9 and a vertical leg of length : :. _ d (, ) - 1 _ 1 ( 1, 1 ) (, 1 ) Since we have a right triangle, we can appl the Pthagorean Theorem, which states that the sum of the squares of its sides is equal to the square of the hpotenuse. Letting d represent the distance between (9, : ) and (9, : ), we have! = : : We are squaring the quantities, thus we can replace the absolute value bars with parentheses.! = (9 9 ) +(: : ) 9

10 Taking the principal square root, we now summarize the distance formula. Distance Formula The distance between points (9, : ) and (9, : ) in the coordinate plane is given b! = 9 9 : : Eample 1 Find the distance between the points (5, 7) and (3, ). Using the distance formula, we have! = 9 9 : : = 3 5) +[ ( 7)] = ( 8) +(11) = +11 = Hence, the eact distance between the given points is 185 units, or approimatel 13. units. Eample The circle below has diameter with endpoints at points P 1 and P. Calculate the diameter of this circle. P (1., 1.) P 1 (-1., -1.) Using the distance formula, we have! = [(1. ( 1.)] +[1. ( 1.)] = (.) +(3.) = = 1 = The diameter of the given circle has a length of units. 10

11 Midpoint Formula As the name suggests, the midpoint of a line segment is the point that lies halfwa between the endpoints of the line segment. If we have two points a and b on a number line, the midpoint, M, is located eactl midwa between the two points, and we onl need to average the coordinates, as shown below. = = > &? a To find the coordinates =(9,: of the midpoint of the line segment that connects two points (9, : ) and (9, : ) on the coordinate plane, the formula will be similar, but now we must average both coordinates. M b M(, ) (, ) ( 1, 1 ) Since = lies eactl halfwa between (9, : ) and (9, : ), we know that the distance between 1 and and the distance between and is the same. Therefore, we know that 1 =. Solving for, we have 1 = = In the same wa, : = B A & B +. So, the location halfwa between 1 and is 9 A &@ +, and between 1 and is : = B A & B +. Midpoint Formula The midpoint = = (9,: of the line segment with endpoints (9, : ) and (9, : ) in the coordinate plane is given b = = # 9 9, : +: $ Take the average of the two -coordinates, and the average of the two -coordinates.! CAUTION These are two things to keep in mind: When using the distance formula, we subtract the respective coordinates; since we are finding a distance, the answer is number. When using the midpoint formula, we add the respective coordinates; since we are looking for coordinates of a point, the answer is an ordered pair. 11

12 Eample 3 a. Find the midpoint of the line segment with endpoints (, 7) and (, 5). b. Prove that the distance between (, 7) and the midpoint equals the distance between (, 5) and the midpoint. c. Find half the distance between both endpoints. a. Using the midpoint formula, we have ) & (,, (5& * = ) (0,( * = (, 1) b. Distance between (, 7) and (, 1) = ) +[ 1 ( 7)] = 1+3 = 5 = 13 = 13 Distance between (, 5) and (, 1) = [( ( )] +( 1 5) Both distances are equal. = 1+3 = 13 c. [Distance between (, 7) and (, 5)] = [(5 ( 7)] +( ) = 1+ = 08 = 1 13 = 13 = 13 Observe that the distance between the midpoint and each of the endpoints is equivalent to half the distance between both endpoints. Eample The graph below shows the approimate total revenue (in millions) for Google between the ears 007 and 011. Assuming that the compan's performance followed a linear pattern, use the midpoint formula to estimate its revenue for 009. Source: Google Total Revenue ($ millions) 1,59 37,905 Year

13 Using the midpoint formula, we have )..5 &.,0 & 5., * = (009, 79.5) The estimate revenue for 009 is $7,9.5 (in millions). Note: The actual revenue reported b Google for 009 was $3,51 (in millions); the eample assumes a linear pattern, which not necessaril reflects the actual compan's performance trend. Circles How do we define a circle? A circle is the set of all points (9,: in the coordinate plane that are equidistant from a fied point, (h,d, called the center. The fied distance between the circle's center to an point (9,: is called the radius, E, where E > 0. r (, ) (h, k) We can use the distance formula to find the standard form of the equation of a circle. Let (h,d represent the coordinates of the center of a circle, and assume that (9,:is an point on the graph of the circle. The distance between the points h,d and (9,: is equal to the radius, E. Appling the distance formula, we have 9 h : D = E This gives us the standard form of the equation of a circle. 9 h : D = E Square both sides. The Standard Form of the Equation of a Circle For h, D, and E real numbers, E > 0, the general form of the equation of the circle with center (h,d and radius E is given b 9 h : D = E The equation of the circle with center at (0,0) and radius E is given b 9 +: = E Eample 5 Find the standard form of the equation of each circle, given its center and radius. a. Center (5, ); radius 3 b. Center (, 0); radius 5. c. Center (0, 0); radius 1 13

14 a. We will use 9 h : D = E with (h,d = (5, ) and E = [: )] = 3 (9 5) +(: +) = 9 b. Since (h,d) = (, 0) and E = 5, we have (9 ) +: = ( 5) (9 ) +: = 5 c. The center is located at the origin, thus we will use 9 +: = E with E = : = 1 9 +: = 1 Note: The circle with center at (0, 0) and radius E = 1, 9 + : = 1, is called the "unit circle" and it is used frequentl in trigonometr. Eample Find the standard form of the equation of a circle with center at (, 3), which passes through the point (3, 1). We know that the radius E of the circle is the distance between (, 3) and (3, 1). E = (3 ) + [1 ( 3)] = (1) +() = 1+1 = 17 Since (h,d) = (, 3) and r = 17, the equation of the circle is Graphing Circles (9 ) +[: ( 3)] = F 17 G (9 ) +(: +3) = 17 Given the standard form of the equation of a circle, we can identif the center and the radius to help us graph the circle. Once we plot the center point, we count E units left and right from the center, and up and down from the center to plot four ke points. Then, we connect the resulting points with a smooth curve to sketch the circle. The steps to graph the standard form of the equation of a circle are summarized net. Graphing a Circle of the Form (H I) J +(K L) J = M J (1) Identif the center (h,d) and the radius E. () Plot the center point. (3) Starting from (h,d), count E units horizontall (left and right from the center) and verticall (up and down from the center) to plot four ke points. (3) Connect the four ke points with a smooth curve to sketch the circle. () Label the center and the radius. 1

15 Eample 7 a. Graph the circle represented b 9 5 : ) = 9. Label the center and the radius. b. Find the intercepts, if an, of the graph of this circle. a. The standard form of the equation of a circle gives its center and radius. Rewriting the given equation in the form (9 5) + [: ( )] = 3, we know that the center (h,d) = (5, ), and the radius E = 3. Plotting the center, counting 3 units verticall and horizontall, and completing the circular sketch, we obtain the graph that follows. r = 3 (, ) (5, -) b. To find an 9-intercepts, we let = 0 and solve for : (9 5) +(0+) = 9 (9 5) + = 9 (9 5) = = ± 5 9 = 5 ± 5 The -intercepts are 9 = and 9 = c. To find an -intercepts, we let = 0 and solve for : (0 5) +(: +) = 9 5+(: +) = 9 (: +3) = 1 Since (: +3) cannot be negative, this circle does not have an :-intercept, as shown net. 9-intercepts (.7, 0) (7.3, 0) 15

16 General Form of the Equation of a Circle Suppose we have the circle 9 3) +(: +1) = 5. Epanding this equation, we have : +: +1 = 5 Square each binomial : +: +1 5 = 0 Subtract 5 from both sides. 9 + : 9 +: 15 = 0 Combine constants and rearrange terms. The resulting equation is the equation of the circle written in general form. The General Form of the Equation of a Circle For A, B, C, D, and E real numbers, A = B, A and B not zero, the general form of the equation of the circle is given b P9 + Q: + R9 +S: +T = 0 Notice that while the standard form of the equation of the circle eplicitl gives us information about its center and radius, the general form does not. When given the general form of the equation of a circle, we can rewrite the equation in standard form b using the process of completing the square. Eample 8 Write each equation in standard form. Find the center and radius of each circle. a. 9 + : 9 10: 10 = 0 b. 9 + : 9 +: = 1 a. We start b grouping an epressions involving the 9 variable, those involving the : variable, and adding 10 to both sides. 9 + : 9 10: 10 = 0 (9 9) +(: 10:) = 10 Net, we need to complete the square on both 9 and :. To complete the square on 9, square half the coefficient of 9 and add this square to both sides of the equation. Similarl, to complete the square on :, square half the coefficient of : and add this square to both sides of the equation. (9 9 +1) + (: 10: +5) = U ( )V = 1 U ( 10)V = 5 Add 1 to both sides Add 5 to both sides Now we factor each trinomial, add the constants on the right-hand side of the equation, and write the equation in standard form. (9 1) + (: 5) = 3, or (9 1) + (: 5) = From the standard form of the equation we can see that the center of this circle is (1, 5) and its radius is. 1

17 b. To complete the square, the coefficients of 9 and : must be 1.Therefore, dividing both sides of the equation 9 : 9 : = 1 b, we obtain 9 + : 9 +: = 0 We can now rearrange and regroup the terms, and complete the square on both 9 and :. (9 9) + ( : +:) = 0 ) * +): +: + 0 * = Rearrange and regroup terms. Complete the square on both and. U ( 1)V = 0 Add to both sides Add 0 U (1)V = 0 to both sides 0 )9 * + ): + * = 0 Factor and add the constants. )9 * + ): + * = ) * So, the center of this circle is ), * and its radius is 0.87.! CAUTION Although we can write ever equation of a circle in the general form, not ever equation of this form will represent the equation of a circle. For eample, 9 + : 9 10: + 7 = 0 is equivalent to 9 + : 9 10: = 7. Completing the square will result in (9 1) + (: 5) = 1, which gives us E < 0. Thus, there are no real numbers 9 and : that satisf this equation, and it has no graph. Additionall, let us consider the equation 9 + : 9 10: =. Completing the square, we obtain (9 1) + (: 5) = 0. Since E = 0, this equation would not represent a circle, but rather a single point. Eample 9 The London Ee Ferris wheel has a diameter of 135 meters. Passengers get on the capsules on the lowest point of the wheel from a platform meters above ground level. Assume that we establish a coordinate plane with the 9-ais running across ground level, and the center of the wheel located on the :-ais. Find the equation in standard form for the circular wheel. Little Facts: The London Ee Ferris wheel is located on London's South Bank, across the Westminster Bridge. It opened to the public in 000 and, as of 01, it was considered one of the most visited attractions in London, with approimatel 3.5 million visitors per ear. Interestingl, there are 3 passenger capsules on the wheel, but for superstitious reasons the are numbered 1-33; there is no capsule #13. Sources: The center of the wheel is located on the :-ais, which means that its 9-coordinate is 0. Since the diameter is 135 meters, we know that the radius is 135 = 7.5 meters. Given that the lowest part of the wheel comes within meters of the ground, this means that the :-coordinate of the center will be at = 9.5 meters. Therefore, the center of the wheel has coordinates 17

18 (0, 9.5). Thus, the equation of the circular wheel is (9 0 : 9.5 = 7.5, or equivalentl, 9 :9.5 = Graphing Calculator and Circles To graph the equation of the circle using the grapher, we must first isolate the :. Let us graph the equation 9 : =. Solving for : in terms of 9 we have : = 9 : = N 9 : = N 9 We let Y1 = 9 to graph one-half of the circle, and Y = 9 to graph the other half. Since the default viewing window of the grapher is not a square, it is best use the ZSquare window (#, option 5) so that the graph does not look distorted. The graph is shown net. [.7,.7, 1] b [3.1, 3.1, 1] In Eample 7, we sketched the graph of 95 : = 9. Solving for : in terms of 9, we let Y1 = 9 95, and for the other half we let Y = The graph follows. [5, 11.5, 1] b [8.75,.375, 1] Note: When we graph circles using the calculator, man times the circles will not appear completel closed because of the resolution of the calculator. In-Class Practice Section (B) 1. Find the distance between each pair of points: a. (8, 5) and (3, 11) b. (3.5, ) and (5,.). Two contestants in a realit show have to travel an undetermined number of miles to an unknown destination. The are onl given the following information: Starting at coordinates (15, 0), the first da the need to travel to coordinates (10, 80); the second da the will travel west to a resting point with -coordinate 5. What was the total distance the traveled? Round our answer to decimal places. 3. Find the midpoint of each line segment with the given endpoints. a. (10, 8) and (, ) b. ),* and 5 ), * c. 7, and 7, 0 18,

19 . a. Find the standard form of the equation of a circle with center (, 1) and radius. b. Graph the circle and label the center and the radius. c. Find the intercepts, if an, of the graph of this circle. 5. Suppose a merchant ship is located at coordinates (, ) and has a radar sstem with 30 range of 0 nautical miles. Write the equation of the circle that would model the range of the radar. Note: One nautical mile 1.15 miles.. Write the equation 9 : 89: = 0 in standard form and identif the center and radius of the circle. Eercises Section (B) In eercises 1-, find the distance between the given pair of points. Write our answers in radical form, then round to decimal places. 1. (, 7) and (, 11). ( 3, ) and ( 5, ) 3. ( 9, 8) and ( 1, ). (5, 7) and (, 3) 5. ),* and ) 1, *. ), * and ) 5 0, * In eercises 7-1, find the midpoint of each line segment with the given endpoints. Where needed write answer(s) in fraction form. 7. (, ) and (, 9) 8. (10, 1) and (, ) 9. ( 9, 3) and (3, 5) 10. (, 11) and ( 5, 7) 11. ) 5,* and ), * 1. ) 5, 8* and ), *, In eercises 13-18, find the standard form of the equation of a circle with the given center and radius. 13. Center (, 3); radius 1. Center (7, ); radius Center ( 5,); radius 1. Center ( 1, 3); radius Center (, 0); radius Center (.5, 3.8 ); radius.7 In eercises 19-, find the standard form of the equation of a circle with the given center and specified point. 19. Center (, 3); point (,5) 0. Center (, 3); point (,) 1. Center (, 1); point (, 3). Center ( 3, 7); point (,) 3. Center (5, 3); point (1, ). Center (, 7); point (, 10) In eercises 5-9, graph the following circles; label the center and the radius. 5. (9 1) +(: +3) = 5. (9 +3) +(: 5) = 7 7. (9 +) +(: 7) = 3 8. (9 ) +(: +1) = 9. (9 +) +(: +) = 9 In eercises 30-35, do the following: a. Write in standard form. b. State the center. c. State the radius : 9 8: 5 = : +9 1: = 7 19

20 3. 9 1: + : +9 = : : = : 109 +: = : 5: = 0 In eercises 3-1, find the standard form for the given circles (3, -1) (, -) (-3, ) (-1, ) (-5, 1) (3, -1) (7, -3) (-5, 5) (-, 3) (, 3) (3, 1) Lake Eola s fountain in downtown Orlando, Florida, has a radius of 30 feet. The lake is well-known for its swans and swan boats that are popular to locals. The bottom of the lake to the center of the fountain s base is feet. Assume that we establish a coordinate plane with the 9-ais running across the bottom of the lake, and the center of the fountain located on the :-ais. Find the equation in standard form for the circular fountain. Little Facts: Lake Eola was originall a sinkhole, and Lake Eola Park was established in The original sinkhole is north of the fountain with a depth of 80 feet. The first fountain was built in 191, costing $10,000; the second fountain was built in 1957at a cost approimatel $350,000. Source: 3. Rand is a drag racing enthusiast and had to bu new back tires, costing $390 a piece. He measured the radius of the tire to be 15.5 inches, and the width measuring 5.35 inches. 0

21 Assume that we establish a coordinate plane with the 9-ais running across the surface of the tire, and the center of the tire located on the :-ais. Find the equation in standard form for the circular tire. Round all values to decimal places where needed. Little Facts: With a mere.5 to 7.5 psi, top fuel dragster back tires grow approimatel 8.5 inches in diameter when reaching a 35 mph speed run. The dragsters tpicall accelerate from 0 to 100 mph in approimatel just 0.8 seconds. Sources: The graph below shows the approimate opening stock price of Jarden Corporation at the beginning of each month from Januar to November in 01. Assuming that the compan's opening stock price at the beginning of each month followed a linear pattern, use the midpoint formula to estimate the opening stock price in June 01. Round all values to decimal places where needed. Little Fact: Jarden Corporation sells a diverse line of products; their portfolio contains, to name a few, Rawlings, Crock-Pot, Coleman, Oster and Mr. Coffee. Source: stock price 30 5 (11, 33.89) 0 15 (1, 0.1) 10 5 Jan Feb Mar Apr Ma Jun Jul Aug Sep Oct Nov month 5. Dann leaves Akron, Ohio and starts driving to the Rock and Roll Hall of Fame and Museum in Cleveland. His friend Eduardo notices that their place of departure is located at (8, 1) on a map s coordinate grid. The stop halfwa for lunch in Parma, then head towards the museum, whose coordinates are (0, 8). Find the location of Parma on their map. Little Facts: Alan Freed was a disc jocke in Cleveland, Ohio in 1953, in New York he was known as The King of Rock n Rollers with thousands of followers. As he rose to fame, he was considered the Pied Piper that promoted juvenile delinquenc via rock n roll. The Rock and Roll Hall of Fame built on the shore of Lake Erie in 1995, is situated in Cleveland because it is where man historians sa the first rock n roll concert took place in 195, at the Cleveland Arena, featuring The Moondog Coronation Ball that was a stage dance with R&B stars. Sources: Suppose ou situate a rectangular coordinate sstem on a baseball field in which home plate is the origin, home plate to 1 st base is the positive 9-ais and home plate to 3 rd base is the positive :-ais. The distance between each base is 90 feet. Assume a left fielder s coordinate on the field is (85, 30). If the third baseman s coordinate is (, 9), find the distance that the left fielder s ball will travel to 3 rd base. Round our answer to the nearest foot. 7. A golfer is on his second stroke; his first stroke landed 3 feet short and 9 feet to the left of the th hole. His second stroke landed feet past the hole and feet to the right. Assuming that the hole is situated at the origin and the ball s path was linear, find the distance his golf ball traveled on the second stroke. Round our answer to the nearest foot. 1

22 Answer Ke Section (B) Eercises (1, 7/) 9. (3. 1) 11. (1/, 9/5) : 3) = (9 +5) +(: ) = 17. (9 ) +: = 19. (9 +) +(: 3) = 0 1. (9 ) +(: 1) = 3 3. (9 5) +(: +3) = r= (1, -3) (-, 7) r= (-, -) -3 r= a. (9 +3) +(: ) = 5 b. ( 3, ) c a. (9 10) +(: +5) = 19 b. (10, 5) c a = b. (3/, 5/) c (9 +3) +(: ) = (9 7) +(: +3) = 0 1. (9 3) +(: 1) = (: 5.35) = (1, 11) 7. 0 ft.

23 Section C: Snthetic Division Division of Polnomials We know that we can use long division to check whether or not the divisor is a factor of the dividend. Let us review the long division process. Divisor Quotient Dividend Remainder Is 1 a factor of 9? The remainder is 0, thus 1 is a factor of 9. Checking: (1)(1) = 9. Now, is the binomial 9 3 a factor of ? We can use long division of polnomials to answer this question. Polnomial division is ver similar to long division of numbers. A first step is to make sure that both the dividend ( ) and the divisor (9 3) are polnomials written in descending order. Let us review the process of long division of polnomials Divide 9 b 9. So, the first term of the quotient is 9. ( 9 39 ) 9 (9 3) = Subtract; bring down net term Divide 59 b 9. So, the second term of the quotient is 59. (59 159) 59(9 3) = Subtract; bring down net term. 9 3 Divide 9 b 9. So, the last term of the quotient is 1. (9 3) 1(9 3) = 9 3. Subtract. 0 Remainder Since the remainder is 0, 9 3 divides evenl into , and we conclude that 9 3 is a factor of The quotient, , will be the other factor of the dividend. We can check b multipling: (9 3)( ) = Eample 1 Divide b Observe that both the dividend and the divisor are missing first-degree terms. In this case, we insert 0's for the missing 9 terms. The long division follows. 3

24 Divide 99 b 39. ( ) Multipl, subtract, and bring down net term Divide 159 b 39. ( ) Multipl, subtract, and bring down net term Remainder When the remainder is 0 or the divisor has a higher degree than the remainder, we stop the division process. So, the quotient for this division is with a remainder of The solution to a long division problem can be written as YZ[ZY\]Y YZ[Z^_` In our &@ + + & = &. Net, we summarize the long division process. The Division Algorithm = quotient + `\abz]y\` YZ[Z^_`. Let d(9) and!(9) be polnomials with!(9) 0, and degree of!(9) less than the degree of d(9). There eist unique polnomials e(9) and E(9) such that d(9)!(9) = e(9) + E(9)!(9) or d(9) =!(9) e(9) +E(9) Dividend Divisor Quotient Remainder where E(9) = 0, or its degree is less than the degree of!(9). B the division algorithm, we can epress the solution to Eample 1 as follows = (39 + 1)( 39 +5) + ( 39 19) Dividend Divisor Quotient Remainder Snthetic Division Snthetic division is shortcut method of dividing polnomials that can be used when the divisor is a first-degree polnomial of the form 9, where is a constant. Using snthetic division, the process of division is simplified b omitting all the variables, and using onl the coefficients of both dividend and divisor. Of course, we still use 0s for missing powers. Let us divide b 9 5 using long division, and compare it to the snthetic division of the same polnomials.

25 Long Division Snthetic Division For the snthetic division process, we start b omitting all the variables, and using onl the coefficients of both dividend and divisor. The coefficients of the dividend are, 13, and 15. The divisor 9 5 is equivalent to 9 5, hence = 5. Since the coefficient of 9 in the divisor 9 will alwas be 1, we can omit it when we work with snthetic division. We draw a line below the coefficients, leaving room above the line Once we have set up the snthetic division (removing the variables), we bring down the leading coefficient Now we multipl the divisor, 5, b the we brought down, and write this product in the second row, second column. Then, add the values in the second column: (13) + (10) (9 109) Net, multipl the divisor 5 b the 3 on the bottom row in the second column, and write this product in the second row, third column. Then, add the values in the third column: (15) + ( 15) (9 109) (39 +15) 0 The last entr of the final row on the snthetic division will give us the remainder of the division process. The other entries of the final row, will give us the coefficients (in descending order) of the terms of the quotient. The quotient will be a polnomial whose degree is 1 less than the degree of the dividend. In our eample, we have 3 0 Coefficients of the quotient Remainder So, our quotient is 9 3, and we have a remainder of 0. 5

26 Eample Use snthetic division to divide b First, we rewrite the dividend in descending order, and insert 09: The coefficients of the dividend are 3, 5, 0, and 9. Since the divisor is 9 +1, which is equivalent to 9 ( 1), the value of the constant is Set up the snthetic division (remove the variables) and bring down the leading coefficient Multipl the divisor, 1, b the 3 on the bottom row and write the 3 product in the second row, second column. Add the values in the second column: ( 5) + ( 3) Multipl 1 b 8 on the bottom row and write the product in the 3 8 second row, third column. Add the values in this column: Multipl 1 b 8 on the bottom row and write the product in the second row, net column. Add these values: ( 9) + ( 8) The entries 3, 8, and 8 will give us the coefficients (in descending order) of the terms of the quotient, and the remainder is the last entr, 17. So, dividing b 9 +1 results in It is important to remember that the degree of the quotient will be 1 less than the degree of the dividend. Net, is a summar of the snthetic division process.

27 Snthetic Division To divide a polnomial d 9 b a first-degree polnomial of the form 9, where is a constant, we can follow these steps. (1) Write the coefficients of dividend, d 9, in descending order, and insert 0's for an missing terms. () Determine the value of (from the divisor) and set up the snthetic division. (3) Draw a line below the coefficients, leaving room above the line, and bring down the leading coefficient. () Multipl b the coefficient ou brought down, and write this product in the second row, second column. Add the values in the second column. (5) Multipl the divisor,, b the result from step, and write this product in the second row, third column. Add the values in the third column. () Repeat the process until all the coefficients have been used and all columns have been filled in. (7) The last entr of the bottom row is the remainder of the division process. The other entries in this row will give the coefficients (in descending order) of the terms of the quotient, which will be a polnomial whose degree is 1 less than the degree of d 9. With practice, we can abbreviate the process. We will show the shorter version using the problem from Eample Set up the snthetic division format Bring down the leading coefficient, (1)(3) = 3; 5) 3) = (1)(8) = 8; = 8 (1)(8) = 8; 9) 8) = 17 The quotient is and the remainder is 17. The Remainder Theorem Notice that the remainder obtained when we divide a polnomial function d(9) b a first-degree polnomial, 9, is alwas a constant. This occurs because the degree of the remainder must be less than 1 (that is, less than the degree of 9 ). Let us revisit the Division Algorithm, d(9) =!(9) e(9) +E(9). Letting!(9) = 9, and knowing that the remainder E(9) will be a constant, we have d(9) = (9 ) e(9) + E Dividend Divisor Quotient Remainder 7

28 Suppose we evaluate d 9 when 9 =. We get d = e E d = 0) e( ) + E d( ) = E This leads us to a new wa of evaluating polnomials, which we state net. The Remainder Theorem If a polnomial d(9) is divided b 9, then the remainder is d( ). In Eample we used snthetic division to divide d(9) = b 9 +1 and the remainder was 17. Since the divisor is equivalent to 9 ( 1), the value of = 1. Let us now substitute 1 for 9 and evaluate d( 1). d( 1) = 3( 1) 5( 1) 9 d( 1) = = 17 Observe that d( 1) = 17, which is the remainder we obtained via the snthetic division of d(9) b The remainder theorem implies that we can use snthetic division to evaluate a polnomial function d(9) at 9 =. When we divide d(9) b 9 the remainder will be d( ), that is, the value of the polnomial at. Eample 3 Let d(9) = Use the remainder theorem to find d( ). We will use snthetic division with =. The coefficients of the dividend are 5, 3, 1, 1, and B the remainder theorem, d( ) = 119. Remainder Optionall, we can verif our finding b direct evaluation: d( ) = 5( ) 0 3( ) +( ) ( ) +9 = = 119 8

29 The Factor Theorem Earlier, we divided b 9 3 and we obtained the quotient with a remainder of 0. Since 9 3 divides evenl into , we were able to conclude that 9 3 was a factor of the dividend. Checking our answer, we noticed that the multiplication of (9 3)( ) = B the remainder theorem, if d(9) is divided b 9, and the remainder is 0, then we know that d( ) = 0. Net, we summarize the relationship between the first-degree factor, 9, the constant, and d( ) = 0. Factor Theorem If d(9) is a polnomial function, then 9 is a factor of d(9) if and onl if d( ) = 0. This implies that if d( ) = 0 then 9 is a factor of d( ) = 0. Note: The Factor Theorem is revisited in Section 7.1. We can use the remainder theorem and the factor theorem to determine factors of polnomials of higher degrees. Eample Determine if 9 +5 is a factor of the given polnomials. a. d(9) = b. d(9) = a is equivalent to 9 ( 5), and the value of the constant is 5. B the factor theorem, if 9 is a factor of d(9), then d( ) = d( 5) = 0. Using snthetic division, we have B the remainder theorem, d( 5) = 0. Therefore, we know that 9 +5 is a factor of d(9) = b. Using snthetic division, B the remainder theorem, d( 5)= 19. Since the remainder is not 0, we conclude that 9 +5 is not a factor of (9) = Eample 5 Given that 9 + is a factor of d(9) = , find all possible real zeros. Using snthetic division, we have the following: 9

30 The quotient is and the remainder is 0, hence we can write the given polnomial as (9 +)( ). We can factor the quotient as (9 1 )(9 1). Thus, the factorization of d(9) = is given b d(9) = (9 +)(9 1 )(9 1). The zeros can be found when d(9) = (9 +)(9 1 )(9 1) = 0. So, the real zeros of the given polnomial are, and 1. Note: if 9 + is a factor of a polnomial, is a zero. In general, if 9 + is a factor of a polnomial, is a zero of that polnomial. Eample The attendance, P, (in millions) to Magic Kingdom, in Orlando, FL, for the ears 005 to 011 can be modeled b P(9) = , where 9 represents the number of ears since 005. Comparing the ears 008 and 010, which had a higher attendance? Use the remainder theorem to answer the question. Sources: Since 9 represents the number of ears since 005, 9 = 3 represents 008, and 9 = 5 is 010. First, we will find P(3). Net, we will find P(5) P(3) = P(5) = 1.87 The attendance (in millions) in 008 was 17.1, and in 010 it was Therefore, 008 had more visitors. Note: The actual attendance (in millions) reported b Magic Kingdom for 008 and 010 was 17.0 and 1.9, respectivel, which is ver close to the approimation provided b our model. In-Class Practice Section (C) 1. Use long division to divide: ( ) (89 + ).. Use snthetic division to divide: ( ) (9 +1). 3. Let d(9) = Use the remainder theorem and snthetic division to find d( 3). 30

31 . Determine if 9 is a factor of the given polnomials. a. d 9 = b. d(9) = Given that 9 is a factor of d(9) = , find all possible real zeros.. The attendance, P, (in millions) to Disneland, in Anaheim, CA, for the ears 005 to 011 can be modeled b P(9) = , where 9 represents the number of ears since 005. Comparing the ears 007 and 011, which had a higher attendance? Use the remainder theorem to answer the question. Sources: Eercises Section (C) In eercises 1-, divide using long division. State the quotient and the remainder ( + 0) ( + ). ( 3) ( ) 3. ( + ) ( 1). ( ) ( ) ( ) ( ). ( ) ( 3) In eercises 7-11, divide using snthetic division. State the quotient and the remainder ( + 3 1) ( ) 8. ( 9 + 5) ( + 3) ( 7 + 1) ( + 1) 10. (3 9) ( ) 11. ( + 30 ) ( + ) In eercises 1-15, given the following polnomials use the remainder theorem and snthetic division to find the indicated function value f ( ) = ; f ( 1) 13. g ( ) = ; g () 3 1. h ( ) = ; h () 15. f ( ) + 3 = ; f ( ) 1. Determine if + 3is a factor of the given polnomials. a. f 3 ( ) = + 3 b. g ( ) = c. p ( ) = In eercises 17-, find all possible real zeros of each polnomial, given its factor. Write answers in fraction form where needed p ( ) = ; ( +1) 18. f ( ) = ; ( + 7) f ( ) = ; ( ) 0. h ( ) = ; ( + 3) g ( ) = ; ( ). f ( ) = ; ( 5) 31

32 3. The attendance, P, (in thousands) to a baseball pitching tournament for the ears 1990 to 010 can be modeled b P 9 = , where 9 represents the number of ears since Comparing the ears 007 and 008, which ear had higher attendance? Round the attendance number to decimal places. Use the remainder theorem to answer the question. Little Facts: The rotation of the knuckleball is vastl different from the other 3 main pitches, that is, the curveball, fastball, and slider. A tpical fastball will make around 1 revolutions, whereas a tpical knuckleball will make between 1.5 to.5. The reason a knuckleball is so difficult to pitch is that it is more pushed rather than thrown. Sources: Sall owns a gummi bear distribution compan that specializes in all natural flavored gummies. The sales of her compan, f, (in thousands) for the ears 000 to 01 can be modeled b f(9) = , where 9 represents the number of ears since 000. Comparing the ears 005 and 011, which ear had higher sales? Round the sales to decimal places. Use the remainder theorem to answer the question. Little Facts: German Hans Riegel invented the first gummi bears in the 190 s. He soon formed the compan Haribo, which stands for Hans Riegel Bonn. The first Haribo gummi bear was sold in the US in 198. Sources: Answer Ke Section (C) Eercises ; R : ; R : ; R : ; R : ; R : , 0, 1 19.,0, 1. 0,, ; 1.8 (in thousands) 0 3

33 Section D: More on Rational Functions In section 7., we learned about vertical and horizontal asmptotes of rational functions. Let us revisit the aforementioned with an eample. Eample 1 Given the rational function: d 9 ( a. Find an horizontal or vertical asmptotes. b. Graph the function along with an horizontal or vertical asmptote(s). a. Function d is alread in lowest terms. Horizontal asmptote: Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asmptote. Vertical asmptote: Solving 9 = 0, we obtain 9 =. b. We alread determined the asmptote. Following the steps delineated in Section 7., we have The domain of the function is the set of all real numbers ecept, or (, (,. d(0 = (.+ ( = ( = 1, thus the -intercept is (0,1)..( ( Setting the numerator 9 = 0 and solving for, we get = ±. The -intercepts or zeros for the given function are ( 0) ( 1.1,0 and ( 0) (1.1,0. The following table of values contains additional points to helps us sketch the graph: The graph is displaed below. 1 3 f() f ( ) = = - Oblique (Slant) Asmptotes Some graphs of rational functions have another tpe of asmptote, which is neither horizontal nor vertical. When the degree of the numerator is eactl one more than the degree of the denominator, an oblique asmptote occurs. This asmptote also gives us information about the behavior of the graph as (that is, as approaches negative infinit) and as (as approaches positive infinit). 33

34 Note: Some authors prefer to use the word slant rather than oblique. Both terms ma be used snonmousl. We will use the term oblique throughout this topic. Let us consider the function from Eample 1. To find an oblique asmptote, we divide the numerator, 9, b the denominator, 9. Since we have a linear divisor of the form 9, we can use snthetic division with =. The coefficients of the dividend are 1, 0, and The quotient is 9 + and the remainder is. Thus, d(9 = Observe that as or, the (. approaches 0, and we see that the graph of the function d approaches the graph of : = 9, So, we sa that the line : = 9 + is an oblique asmptote of the graph of d. The graph from Eample 1, along with the vertical and oblique asmptotes, is shown below. f ( ) = = = + Oblique (Slant) Asmptote of a Rational Function Let d(9 = i(@ be a rational function in lowest form, such that the degree of k(9 is j(@ eactl one more than the degree of!(9. The graph of d will have an oblique asmptote. To find the oblique asmptote, divide k(9 b!(9, where the quotient is e(9 and the remainder is E(9 d(9 = e(9 + (@ j(@ Since the degree of k(9 is one more than the degree of!(9, the quotient e(9 is a linear function, which will be the oblique asmptote of d. The remainder becomes almost zero, and we discard it. Following is a summar of the different linear asmptotes of rational functions. 3

35 Summar of Linear Asmptotes of Rational Functions Let d 9 = be a rational function such that k 9 and! 9 have no common factors. Vertical Asmptotes: The line 9 = D is a vertical asmptote of graph of the function if! D = 0. A rational function ma have man vertical asmptotes or none at all; the graph of d will never intersect a vertical asmptote. Horizontal Asmptotes: The line : = D is a horizontal asmptote of the graph of the function if d 9 approaches D as 9 or as 9. Let d 9 = = > l@ l &> lma & & > + &> > o? p &? pma & &? + &? o, where q 0, 3 0, r = degree of the numerator, and s = degree of the denominator. If r < s, the line : = 0 (that is, the 9-ais) is the horizontal asmptote of d. If r = s, the line : = > l? p (ratio of the leading coefficients) is the horizontal asmptote of d. If r > s, the function does not have a horizontal asmptote. Oblique Asmptotes: If the degree of k 9 is eactl one more than the degree of! 9, the graph of d will have an oblique asmptote. Divide the numerator b the denominator and discard the remainder. The quotient is a linear equation, : = r9, which will be the oblique asmptote of d. A rational function ma have at most one horizontal or one oblique asmptote, but never both; the graph of d ma intersect its horizontal or its oblique asmptote. Note: Remember that an asmptote is not part of the graph of the function. In section 7., we eamined properties and graphs of rational functions, including an corresponding vertical and horizontal asmptotes, and removable discontinuities or holes. We also saw their applications in different real-world scenarios. In this section, we have alread amplified our knowledge of asmptotes, with rational functions that contain oblique asmptotes. We will now epand our method for graphing rational functions. Net, is a summar of the strategies we can appl to sketch the graph of a rational function. 35