MAC 1105 Final Exam Review And Practice Final Exam Solutions • You will need to have your own graphing calculator for the test. • You may not share calculators or use any type of communication device in place of a calculator. • The final exam is comprehensive over the entire course (not just the second half). • Students who do not take the final exam will receive a zero for the final exam unless prior arrangements have been made for a rare incomplete. • If you miss the final and do not email me by the day of the exam, I will assume you wish to take a zero on your final exam. Communication with me via email is key! How to Study for the Final Exam • Complete all assigned homework. Remember that an overall (for the entire semester) homework score of at least 80% gives you 10 bonus points on the test. • Review your notes and applicable worksheet questions for each of the 25 objectives in the chart below. • Set up a study system (note cards for example) for each of the 25 objectives in the chart below. See the column below labeled “important ideas”. • Complete the practice test and check your solutions in class. • Work the suggested text problems for each of the 25 objectives in the chart below and check your solutions. The problems are odd numbered so answers should be readily available. Exam Topics 1 – 5 Objective Sections 1) Solve a quadratic equation using the quadratic formula. 1.4 2) Use the discriminant to determine the number and type of solutions of a quadratic equation. 1.4 3) Solve equations graphically. 1.5 Suggested Text Problems Important Ideas p. 165: 31, 33, 35 π₯= −π ± π 2 − 4ππ 2π π· = π 2 − 4ππ p. 165: 47, 49 D = 0 → one real root D > 0 → two real roots D < 0 → two imaginary roots See Notes Enter each equation. (Y = ) Begin with the standard window. (ZOOM 6) Adjust the window to see the intersection (WINDOW) Find the intersection (2nd TRACE (CALC) 5) The x coordinate is the solution. 4) Solve an absolute value inequality. 1.7 5) Given two points, find the distance between the two points and the midpoint. 2.1 p. 166: 99, 101 p. 293: 9a, 11a, 13a |u| < a means |u| ≤ a means |u| > a means |u| ≥ a means −a < u < a −a ≤ u ≤ a u> a or u < −a u≥ a or u ≤ −a π= (π₯2 − π₯1 )2 + (π¦2 − π¦1 )2 π₯1 + π₯2 π¦1 + π¦2 π=( , ) 2 2 Exam Topics 6 – 10 6) Test an equation for symmetry. 2.2 p. 293: 25, 27, 29, 31, 33 (symmetry only) 1) Substitute: x−axis symmetry: (−y) for y y−axis symmetry: (−x) for x origin symmetry: (−x) for x and (−y) for y 2) Determine if the result is the original equation. 7) Find the center and radius of a circle. 2.2 p. 189: 57, 59, 71 8) Write the equation of a circle. 2.2 p. 293: 35, 37 9) Find the slope of a line. 2.3 p. 293: 9c, 11c, 13c 10) Write a linear equation given two ordered pairs of data. 2.3 p. 295: 105a Complete the square (if necessary) to transform the equation to π₯ − β 2 + π¦ − π 2 = π2 ππππ‘ππ β, π , πππππ’π π ππππ‘ππ β, π , πππππ’π π π₯ − β 2 + π¦ − π 2 = π2 πβππππ ππ π¦ m =πβππππ ππ π₯ = π¦2 − π¦1 π₯2 − π₯1 1) Find the slope. 2) Use the point slope formula: y – y1 = m(x– x1) Exam Topics 11 – 15 11) Solve problems involving linear regression. 2.3 p. 295: 111a, 111c Enter the data (STAT EDIT 1) Find the regression coefficients. (STAT CALC 4) Write the equation of the line. (π¦ = ππ₯ + π) 12) Solve a system of two equations by substitution or elimination 5.1 p. 487: 13,15,17,19 13) Work an application of systems of linear equations. 5.4 p. 488: 87, 91,95, 99 Set up a system of two equations in two unknowns. Solve the system. 14) Use your calculator to find intervals of increasing and decreasing. 2.5 p. 236: 11 Graph the function. Adjust the window to show all of the turning points. The intervals of increasing/decreasing are intervals of the domain (x). 15) Classify a function as even, odd, or neither. 2.5 p. 237: 33, 35, 37, 41, 43, 45 1) 2) 3) Substitution: Solve one of the equations for one the variables (usually y). Substitute that result into the other equation. Solve. Elimination: 1) Get the coefficients of one of the variables to be additive inverses if needed (by multiplication). 2) Add the equations to eliminate that variable. 3) Solve. 4) Back substitute to find the value of the other variable. Find −π₯ . π −π₯ = π π₯ → ππ£ππ π −π₯ = −π π₯ → πππ Exam Topics 16 – 20 16) Evaluate a difference quotient. 17) Find function values of a piece-wise function. 18) Write the equation of a function whose graph has been transformed. 2.5 2.6 2.7 p. 239: 79 p. 248: 21a, 47 p. 265: 67, 69, 71, 73 π π₯ + β − π(π₯) β Make sure to pay attention to which “piece” of the function you should use. Be sure to know the ten different transformations from section 2.7: π π₯ ± π → π’π πππ€π π π’πππ‘π (π’π ππ +) π π₯ ± π → ππππ‘ πππβπ‘ π π’πππ‘π (ππππ‘ ππ +) ππ(π₯) → π π‘πππ‘πβππ π£πππ‘ππππππ¦ ππ¦ π ππππ‘ππ ππ π ππ π > 1, 1 π βππ’ππ ππ¦ ππ 0 < π < 1 π π ππ₯ → π βππ’ππ βππππ§πππ‘ππππ¦ πππ π > 1, π π‘πππ‘πβππ βππππ§πππ‘ππππ¦ πππ 0 < π < 1 −π π₯ → πππππππ‘ππ ππππ’π‘ π₯ π −π₯ → πππππππ‘ππ ππππ’π‘ π¦ 19) Find the vertex of a quadratic function. 3.1 p. 307: 53, 55 (vertex only) Use the formula: π π The vertex of π₯ = ππ₯ 2 + ππ₯ + π is − 2π , π − 2π OR Graph the function and find the max/min. 20) Find any vertical or horizontal asymptotes for a given rational function. 3.6 p. 373: 25, 27, 29, 31, 35, 37, 39, 41 Vertical: Simplify the function completely. The vertical line π₯ = π is a vertical asymptote of the function if π causes the denominator to be zero (after simplification). Horizontal: n = degree of numerator, m = degree of denominator π < π → π₯ − ππ₯ππ π = π → π¦ = πππ‘ππ ππ πππππππ πππππ. π > π → ππ π». π΄. Exam Topics 21 – 25 21) Work a variation application. 22) Solve an exponential equation by writing both sides to the same base and/or by taking the (natural/common) logarithm of both sides. 3.8 p. 393: 29, 33, 35, 37 1. 2. 3. 4. 4.1, 4.5 p. 473: 55, 57, 59, 63 Convert both sides to the same base and set the exponents equal to one another (if possible). Use the given information to set up an equation. Solve for k → the constant of variation. Write a new equation using all variables and the constant k. Input the new information and compute the answer. OR Take the (natural/common) logarithm of both sides. Use the third property of logarithms to “move” the variable to the outside. Solve for the variable. 23) Work with compound interest formulas (including compounding continuously). 4.2 24) Write a logarithmic expression in expanded form or condensed form. 4.4 p. 425: 19, 21 π π΄ = π(1 + )ππ‘ π π΄ = ππ ππ‘ p. 453: 19, 21, 23, 25, 39, 43 ππππ ππ = ππππ π + ππππ π π ππππ = ππππ π − ππππ π π ππππ ππ = πππππ π 25) Solve a logarithmic equation by consolidating the logarithms to a single logarithmic expression. 4.3, 4.5 p. 467: 53, 55, 57, 59 1) Isolate all logarithmic expressions on one side of the equation and use logarithmic properties to rewrite the logarithmic expressions as a single logarithmic expression . 2) Rewrite the logarithmic expression as an exponential expression. 3) Solve. 4) Reject values that produce the logarithm of a negative number or the logarithm of zero. Special Note • It is entirely possible that the questions on the test may have additional parts that differ slightly from those in the practice test. • To prepare for these additional parts, be sure you are familiar with your notes, your worksheets, and your homework. • Objectives will remain unchanged. Practice Test Solutions 1a) Solve: π₯ 2 = 5 − 8π₯ π₯ 2 + 8π₯ − 5 = 0 β¨ π = 1, π = 8, π = −5 −π ± π2 − 4ππ π₯= 2π −8 ± 82 − 4(1)(−5) = 2(1) −8 ± 64 + 20 = 2 −8 ± 84 = 2 −8 ± 4 β 21 = 2 −8 ± 2 21 = 2 = −4 ± 21 {−4 − 21, −4 + 21 } 1b) Solve: π₯ 2 + 4π₯ + 8 = 0 π₯ 2 + 4π₯ + 8 = 0 β¨ π = 1, π = 4, π = 8 −π ± π 2 − 4ππ π₯= 2π −4 ± 42 − 4(1)(8) = 2(1) −4 ± 16 − 32 = 2 −4 ± −16 = 2 −4 ± 4π = 2 = −2 ± 2π {−2 + 2π, −2 − 2π } 1c) Solve: π₯ 2 = 15 + 9π₯ π₯ 2 − 9π₯ − 15 = 0 β¨ π = 1, π = −9, π = −15 −π ± π2 − 4ππ π₯= 2π −(−9) ± (−9)2 − 4(1)(−15) = 2(1) 9 ± 81 + 60 = 2 9 ± 141 = 2 9 − 141 9 + 141 { , } 2 2 2) Find the discriminant and the number and type of roots: 8π¦ 2 = −5π¦ − 3 8π¦ 2 + 5π¦ + 3 = 0 β¨ π = 8, π = 5, π = 3 π· = π2 − 4ππ = 5 2− 4 8 3 = 25 − 96 = −71 π· ππ πππππ‘ππ£π β¨ 2 πππππππππ¦ ππππ‘π π· = −71, two imaginary roots 3) Use the intersection – of – graphs method to approximate the solution to nearest hundredth: −3.5 20 + 7 π₯ + 6.4(14ππ₯ + π1 = −3.5 20 + 7 π₯ + 6.4 14ππ₯ + 0.6 π2 = −5 π΄πππ’π π‘ π‘βπ π₯ π‘π −2,2 π‘βππ πβπππ π 2ππ πππππ πΆπππ 5 πβπ π πππ’π‘πππ ππ π‘βπ π₯ − πππππππππ‘π ππ π‘βπ πππππ‘ ππ πππ‘πππ πππ‘πππ. {0.22} 4a) Solve: 8π₯ − 8 ≥ 9 8π₯ − 8 ≤ −9 ππ 8π₯ − 8 ≥ 9 8π₯ ≤ −1 ππ 8π₯ ≥ 17 1 17 π₯ ≤ − ππ π₯ ≥ 8 8 1 17 (−∞, − ]π [ , ∞) 8 8 4b) Solve: 9π₯ + 3 < 19 −19 < 9π₯ + 3 < 19 −22 < 9π₯ < 16 22 16 − <π₯< 9 9 22 16 (− , ) 9 9 5a) Find the distance between P and Q. π 4, −4 , π −4, 2 π= (π₯2 − π₯1 )2 + (π¦2 − π¦1 )2 = ((−4) − 4)2 + (2 − (−4))2 = −8 2 + 6 2 10 = 100 = 10 5b) Find the midpoint of P and Q. π 4, −4 , π −4, 2 π₯1 + π₯2 π¦1 + π¦2 π=( , ) 2 2 4 + −4 −4 + 2 =( , ) 2 2 0 2 = ,− 2 2 (0, −1) 6a) Test the equation for symmetry: π¦ = −2π₯ 3 + 5π₯ π₯: −π¦ = −2π₯ 3 + 5π₯ β¨ π¦ = 2π₯ 3 − 5π₯ ππ π¦: π¦ = −2 −π₯ 3 + 5 −π₯ β¨ π¦ = 2π₯ 3 − 5π₯ ππ ππππππ: −π¦ = −2 −π₯ 3 + 5 −π₯ β¨ −π¦ = 2π₯ 3 − 5π₯ β¨ π¦ = −2π₯ 3 + 5π₯ π¦ππ ππππππ 6b) Test the equation for symmetry: π¦ = 2π₯ 2 − 2 π₯: −π¦ = 2π₯ 2 − 2 β¨ π¦ = −2π₯ 2 − 2 ππ π¦: π¦ = 2 −π₯ 2 − 2 β¨ π¦ = 2π₯ 2 − 2 π¦ππ ππππππ: −π¦ = 2 −π₯ 2 − 2 β¨ −π¦ = 2π₯ 2 − 2 ππ π¦ − ππ₯ππ 7) Find the center and radius of the circle. π₯ 2 + π¦ 2 + 8π₯ + 10π¦ = 23 π₯ 2 + 8π₯ + + π¦ 2 + 10π¦ + 1 8 = 4, 42 = 16 2 1 10 = 5, 52 = 25 2 = 23 π₯ 2 + 8π₯ + 16 + π¦ 2 + 10π¦ + 25 = 23 + 16 + 25 π₯+4 π₯ −β 2 + π¦ + 5 2 = 64 2 + π¦ − π 2 = π2 −4, −5 , π = 8 8) Find the equation of the circle that satisfies the given conditions: Cπππ‘ππ ππ‘ 0, −7 , πππππ’π 9 π₯ −β π₯−0 2 2 + π¦ −π 2 = π2 + π¦— −7 2 = 92 π₯2 + π¦ + 7 2 = 81 9a) Find the slope of the line through the given pair of points. −9, −1 , (4, 7) π¦2 − π= π₯2 − 7 − (−1) = 4 − (−9) 8 = 13 8 13 π¦1 π₯1 9b) Find the slope of the line through the given pair of points. −7, −1 , 8, −1 π¦2 − π= π₯2 − −1 − −1 = 8 − −7 0 = 15 =0 0 π¦1 π₯1 10) After three seconds, a bottle has 15 ounces of water. After 10 seconds, the bottle has 43 ounces of water. Write a linear equation that models the amount of water in the bottle in terms of x. 3 π ππππππ β¨ 15 ππ’ππππ : 3, 15 10 π ππππππ β¨ 43 ππ’ππππ : 10, 43 43 − 15 28 π= = =4 10 − 3 7 π¦ − 15 = 4 π₯ − 3 π¦ − 15 = 4π₯ − 12 π¦ = 4π₯ + 3 CHECK: π¦ = 4 3 + 3 = 12 π¦ππ , π¦ = 4 10 + 3 = 43 (π¦ππ ) π¦ = 4π₯ + 3 11) The paired data consists of the costs of advertising (in thousands of dollars) and the number of products sold (in thousands). Find the equation of the regression line that models the data. Cost 9 2 3 4 2 5 9 10 Number 85 52 55 68 67 86 83 73 πππππ π‘βπ "Costs" ππ πΏ1 πππ π‘βπ "ππ’πππππ " ππ πΏ2 πππ΄π β¨ πΆπ΄πΏπΆ β¨ 4 π ≈ 2.79, π ≈ 55.8 π¦ = 2.79π₯ + 55.79 12) Solve the system of equations. 6π₯ + 9π¦ = −69 3π₯ − 2π¦ = −2 6π₯ + 9π¦ = −69 −2 3π₯ − 2π¦ = −2 6π₯ + 9π¦ = −69 −6π₯ + 4π¦ = 4 13π¦ = −65 π¦ = −5 6π₯ + 9 −5 = −69 6π₯ − 45 = −69 6π₯ = −24 π₯ = −4 { −4, −5 } 13) Jim wants to plan a meal with 127 grams of carbohydrates and 1150 calories. If green beans have 7 grams of carbohydrates and 30 calories per half – cup serving and if fried shrimp have 9 grams of carbohydrates and 190 calories per three – ounce serving, how many servings of green beans and shrimp should he use? 1 π₯ = ππ’ππππ ππ ππ’π π πππ£ππππ ππ πππππ πππππ ,π¦ = ππ’ππππ ππ π‘βπππ ππ’πππ βπππππππ ππ π βππππ 2 7π₯ + 9π¦ = 127 30π₯ + 190π¦ = 1150 ππππππππ‘πππ −30 7π₯ + 9π¦ = 127 7 30π₯ + 190π¦ = 1150 −210π₯ − 270π¦ = −3810 210π₯ + 1330π¦ = 8050 1060π¦ = 4240 π¦=4 7π₯ + 9 4 = 127 7π₯ + 36 = 127 7π₯ = 91 π₯ = 13 13 βπππ ππ’ππ ππ πππππ πππ 4 π‘βπππ ππ’πππ βπππππππ ππ π βππππ 14) By using a graphing calculator, find the increasing interval(s) for the function f x = 2π₯ 3 − 3π₯ 2 − 36π₯ + 30 Choose windows large enough to allow you to see the turning points: π₯: −10, 10 π¦: −200, 200 −∞, −2 , (3, ∞) 15a) Determine if the given function is even, odd, or neither. π π₯ = 4π₯ 2 − 2 π −π₯ = 4 −π₯ 2 − 2 = 4π₯ 2 − 2 πΈπ£ππ 15b) Determine if the given function is even, odd, or neither. π π₯ = −5π₯ 5 + 9π₯ 3 π −π₯ = −5 −π₯ 5 + 9 −π₯ = −5 −π₯ 5 + 9 −π₯ 3 = 5π₯ 5 − 9π₯ 3 = −1 −5π₯ 5 + 9π₯ 3 πππ 3 15c) Determine if the given function is even, odd, or neither. π π₯ = 9π₯ 4 − 4π₯ − 6 π −π₯ = 9 −π₯ 4 − 4 −π₯ − 6 = 9π₯ 4 + 4π₯ − 6 ππππ‘βππ 15d) Determine if the given function is even, odd, or neither. π₯ π π₯ = 3 π₯ − 7π₯ −π₯ π −π₯ = −π₯ 3 − 7(−π₯) −π₯ = −π₯ 3 + 7π₯ −1 π₯ = −1 π₯ 3 − 7π₯ π₯ = 3 π₯ − 7π₯ Even π π₯+β −π π₯ 16) Determine the difference quotient, β 2 π π₯ = 10π₯ + 13π₯ − 11 , for the function π π₯+β −π π₯ β 10 π₯ + β 2 + 13 π₯ + β − 11 − (10π₯ 2 + 13π₯ − 11) = β 10 π₯ 2 + 2π₯β + β2 + 13π₯ + 13β − 11 − 10π₯ 2 − 13π₯ + 11 = β 10π₯ 2 + 20π₯β + 10β2 + 13π₯ + 13β − 11 − 10π₯ 2 − 13π₯ + 11 = β 20π₯β + 10β2 + 13β = β = 20π₯ + 10β + 13 20π₯ + 13 + 10β 17) In Country X, the average hourly wage in dollars from 1945 to 1995 can be modeled by π π₯ = 0.076 π₯ − 1945 + 0.36 ππ 1945 ≤ π₯ < 1970 0.184 π₯ − 1970 + 3.07 ππ 1970 ≤ π₯ ≤ 1995 Use π to estimate the average hourly wages in 1950, 1970, and 1990. π 1950 = 0.076 1950 − 1945 + 0.36 = 0.74 π 1970 = 0.184 π₯ − 1970 + 3.07 = 3.07 π 1990 = 0.184 1990 − 1970 + 3.07 = 6.75 $0.74, $3.07, $6.75 18) Write an equation for the function whose graph fits the following description. The graph of π π₯ = |π₯| is shifted three units left, reflected in the y-axis and shifted two units down. π βπππ‘ππ π‘βπππ π’πππ‘π ππππ‘: π¦ = π₯ + 3 πππππππ‘ππ ππ π‘βπ π¦ − ππ₯ππ : π¦ = −π₯ + 3 π βπππ‘ππ 2 π’πππ‘π πππ€π: π¦ = −π₯ + 3 − 2 π¦ = −π₯ + 3 − 2 19) Find the vertex of the parabola π¦ = π₯ 2 + 6π₯ + 10. π = 1, π = 6, π = 10 π 6 6 − =− = − = −3 2π 2β1 2 π −3 = −3 2 + 6 −3 + 10 = 9 − 18 + 10 =1 OR graph the function and find the minimum. (−3,1) 20a) Find the vertical asymptotes, if any, of π₯+5 π π₯ = π₯−4 π₯−4=0 β¨π₯=4 π₯=4 20b) Find the horizontal asymptotes, if any, of −4π₯ 2 + 7 β π₯ = 4π₯ 2 − 1 πππππππ πππ πππ’ππ 4 β¨π¦= − 4 β¨ π¦ = −1 π¦ = −1 21) The volume V of a gas varies inversely as the pressure P on it. The volume of a ππ gas is 190 ππ3 under a pressure of 21 ππ2 . What will its volume be under a ππ pressure of 35 2? ππ Assume the temperature is constant. π π= π π 190 = 21 π = 3990 3990 3990 π= β¨π= = 114 π 35 114 ππ3 22a) Solve the equation 8π₯−2 = 644π₯ 8 π₯−2 = 82 4π₯ π₯ − 2 = 2 4π₯ π₯ − 2 = 8π₯ −7π₯ = 2 2 π₯= − 7 2 {− } 7 22b) Solve. Please round your answer in part b to three decimal places. 2π₯ = 52π₯+1 ln 2π₯ = ln 52π₯+1 π₯ ln 2 = 2π₯ + 1 ln 5 π₯ ln 2 = 2π₯ ln 5 + ln 5 π₯ ln 2 − 2π₯ ln 5 = ln 5 π₯ (ln 2 − 2 ln 5) = ln 5 ln 5 π₯= ≈ −0.637 ln 2 − 2 ln 5 {−0.637} 23a) Determine the final value of the given amount: $820 at 8% compounded continuously for 2 years π΄ = ππ ππ‘ π΄ = 820π 0.08 = 820π .16 ≈ 962.28 $962.28 2 23b) Randy invested his inheritance in an account that paid 6.2% interest, compounded continuously. After ten years, he found that he now had $62,978.62. What was the original amount of his inheritance? 62978.62 = ππ .062β10 62978.62 π= π .62 ≈ 33879 $33,879 23c) Find the time required for an initial investment of $10,000 to grow to $18,000 at an annual rate of 6% if the interest is compounded quarterly. π π΄=π 1+ π ππ‘ 0.06 18000 = 10000 1 + 4 1.8 = 1.0154π‘ ln 1.8 = ln 1.0154π‘ ln 1.8 = 4π‘ ln 1.015 ln 1.8 π‘= ≈ 9.87 4 ln 1.015 9.87 π¦ππππ 4π‘ 24a) Write the expression in expanded form. π₯4π§ log 8 π¦ π₯4π§ log 8 π¦ = log π₯ 4 π§ − log π¦ 8 = log π₯ 4 + log π§ − 8 log π¦ = 4 log π₯ + log π§ − 8 log π¦ 4 log π₯ + log π§ − 8 log π¦ 24b) Write the expression in condensed form. 1 ππππ π₯ + 5ππππ π¦ − 2ππππ π₯ 2 1 ππππ π₯ 2 + ππππ π¦ 5 − ππππ π₯ 2 = ππππ π₯ 1/2 π¦ 5 − ππππ π₯ 2 = ππππ = ππππ ππππ 1 π₯ 2π¦5 π₯2 π¦5 π₯ 3/2 π¦5 3 π₯2 ππ ππππ π¦5 π₯3 3 25a) Solve. log 27 1 − π₯ = 1 273 3 1 3 3 = 1−π₯ 3 27 = 1 − π₯ 27 = 1 − π₯ 1 − π₯ = 27 −π₯ = 26 π₯ = −26 Note the lack of domain issues. When you substitute −26 in for π₯, the result is the cube root of 27 which is 3 and certainly greater than zero. {−26} 25b) Solve. 2 log(π₯ +5π₯ + 16) = 1 101 = π₯ 2 + 5π₯ + 16 10 = π₯ 2 + 5π₯ + 16 π₯ 2 + 5π₯ + 16 = 10 π₯ 2 + 5π₯ + 6 = 0 π₯+3 π₯+2 =0 π₯ = −3, −2 Note there are no domain issues. The first step indicates that the solutions will make the expression inside the logarithm equal to 10, which is certainly greater than zero. {−3, −2} 25c) Solve. log π₯ − 9 = 1 − log π₯ log π₯ − 9 + log π₯ = 1 log π₯ − 9 π₯ = 1 101 = π₯ − 9 π₯ 10 = π₯ 2 − 9π₯ π₯ 2 − 9π₯ − 10 = 0 π₯ − 10 π₯ + 1 = 0 π₯ = 10, π₯ = −1 There are domain issues. π₯ ≠ −1 ππ log −1 − 9 ππππ πππ‘ ππ₯ππ π‘ {10} Extra Practice 1) Solve. 5π₯ 2 − 6π₯ = 4π₯ 2 + 6π₯ − 3 1) Solve. 5π₯ 2 − 6π₯ = 4π₯ 2 + 6π₯ − 3 π₯ 2 − 12π₯ + 3 = 0 β¨ π = 1, π = −12, π = 3 π₯= = = −π ± −(−12) ± 12 ± π2 − 4ππ 2π (−12)2 − 4(1)(3) 2(1) 144 − 12 12 ± 132 = 2 2 12 ± 4 β 33 12 ± 2 33 = = 2 2 = 6 ± 33 {6 − 33, 6 + 33 } 2) Find the discriminant and the number and type of roots: 16x 2 − 88x + 121 = 0 2) Find the discriminant and the number and type of roots: 16x 2 − 88x + 121 = 0 π 2 − 4ππ = −88 2 − 4 16 121 = 7744 − 7744 =0 1 ππππ ππππ‘ 3) Use the intersection – of – graphs method to approximate the solution to nearest hundredth: 4 −ππ₯ + 5 = 9 + 0.4π₯ 3) Use the intersection – of – graphs method to approximate the solution to nearest hundredth: 4 −ππ₯ + 5 = 9 + 0.4π₯ π1 = −ππ₯ + 5 4 π2 = 9 + 0.4π₯ {0.92} 4a) Solve. 3π₯ − 3 ≥ 15 4a) Solve. 3π₯ − 3 ≥ 15 3π₯ − 3 ≤ −15 ππ 3π₯ − 3 ≥ 15 3π₯ ≤ −12 ππ 3π₯ ≥ 18 π₯ ≤ −4 ππ π₯ ≥ 6 (−∞, −4]π [6, ∞) 4b)Solve. 3π₯ + 4 ≤ 19 4b) Solve. 3π₯ + 4 ≤ 19 −19 ≤ 3π₯ + 4 ≤ 19 −23 ≤ 3π₯ ≤ 15 23 15 − ≤π₯≤ 3 3 23 [− , 5] 3 5a) Find the distance between P and Q. π −4, 1 , π −7, −9 5a) Find the distance between P and Q. π −4, 1 , π −7, −9 π= = (π₯2 − π₯1 )2 + (π¦2 − π¦1 )2 ( −7 − −4 )2 + ((−9) − (1))2 = −3 = 2 + −10 109 2 5b) Find the midpoint of P and Q. π −4, 1 , π −7, −9 5b) Find the midpoint of P and Q. π −4, 1 , π −7, −9 π₯1 + π₯2 π¦1 + π¦2 π=( , ) 2 2 −4 + −7 1 + −9 =( , ) 2 2 −11 −8 = , 2 2 11 (− , −4) 2 6) Test the equation for symmetry: π¦ = 2π₯ 2 − |π₯| 6) Test the equation for symmetry: π¦ = 2π₯ 2 − |π₯| π₯: −π¦ = 2π₯ 2 − |π₯| ππ π¦: π¦ = 2 −π₯ 2 − |π₯| β¨ π¦ = 2π₯ 2 − |π₯| π¦ππ ππππππ: −π¦ = 2 −π₯ 2 − π₯ β¨ −π¦ = 2π₯ 2 − |π₯| no π¦ − ππ₯ππ 7) Find the center and radius of the circle. π₯ 2 + π¦ 2 − 2π₯ − 2π¦ − 4 = 0 7) Find the center and radius of the circle. π₯ 2 + π¦ 2 − 2π₯ − 2π¦ − 4 = 0 π₯ 2 − 2π₯ + + π¦ 2 − 2π¦ + =4 1 −2 = −1, (−1)2 = 1 2 π₯ 2 − 2π₯ + 1 + π¦ 2 − 2π¦ + 1 = 4 + 1 + 1 π₯−1 2+ π¦−1 2 =6 π₯ − β 2 + π¦ − π 2 = π2 1, 1 , π = 6 8) Find the equation of the circle that satisfies the given conditions: Cπππ‘ππ ππ‘ −2, −3 , πππππ’π 7 8) Find the equation of the circle that satisfies the given conditions: Cπππ‘ππ ππ‘ −2, −3 , πππππ’π 7 π₯ −β π₯ − (−2 2 2 + π¦ −π 2 + π¦— (−3) (π₯ + 2)2 + π¦ + 3 2 = π2 2 = 7 =7 2 9) Find the slope of the line through the given pair of points. 3, −2 , (2, −3) 9) Find the slope of the line through the given pair of points. 3, −2 , (2, −3) π¦2 − π= π₯2 − π¦1 π₯1 −3 − (−2) = 2−3 −1 = −1 =1 10) A manufacturer produces 50 TVs at a cost of $17,500 and 75 TVs at a cost of $21,250. Write a linear equation that models the cost in terms of the number of TVs 10) A manufacturer produces 50 TVs at a cost of $17,500 and 75 TVs at a cost of $21,250. Write a linear equation that models the cost in terms of the number of TVs 50 πππ β¨ $17,500: 50, 17500 75 πππ β¨ $21,250: 75, 21250 21250 − 17500 3750 π= = = 150 75 − 50 25 π¦ − 17500 = 150 π₯ − 50 π¦ − 17500 = 150π₯ − 7500 π¦ = 150π₯ + 10000 CHECK: π¦ = 150 50 + 10000 = 17500 π¦ππ , π¦ = 150 75 + 10000 = 21250 (π¦ππ ) π¦ = 150π₯ + 10000 11) The paired data consists of the number of newspapers (in thousands) that sell at a given price (in nickels) per copy for a college newspaper. Find the equation of the regression line that models the data. Newspapers 1 2 3 4 5 Price 11 8 6 4 3 11) The paired data consists of the number of newspapers (in thousands) that sell at a given price (in nickels) per copy for a college newspaper. Find the equation of the regression line that models the data. Newspapers 1 2 3 4 5 Price 11 8 6 4 3 πππππ π‘βπ "Newspapers" ππ πΏ1 πππ π‘βπ "πππππ" ππ πΏ2 πππ΄π β¨ πΆπ΄πΏπΆ β¨ 4 π = −2, π = 12.4 π¦ = −2π₯ + 12.4 12) Solve. 5π₯ − π¦ = 5 3π₯ + 2π¦ = −10 12) Solve. 5π₯ − π¦ = 5 3π₯ + 2π¦ = −10 2(5π₯ − π¦ = 5) 3π₯ + 2π¦ = −10 10π₯ − 2π¦ = 10 3π₯ + 2π¦ = −10 13π₯ = 0 π₯=0 5 0 −π¦ =5 −π¦ = 5 π¦ = −5 { 0, −5 } 13) Last year Mrs. Garcia invested $50,000. She put part of the money in a real estate venture that paid 7.5% and the rest in a small – business venture that paid 12%. The combined income from the two investments was $5190. How much did she invest in each? 13) Last year Mrs. Garcia invested $50,000. She put part of the money in a real estate venture that paid 7.5% and the rest in a small – business venture that paid 12%. The combined income from the two investments was $5190. How much did she invest in each? π₯ = πππππ¦ ππ ππππ ππ π‘ππ‘π, π¦ = πππππ¦ ππ π ππππ ππ’π ππππ π π₯ + π¦ = 50000 0.075π₯ + 0.12π₯ = 5190 → 75π₯ + 120π¦ = 5190000 ππππππππ‘πππ −75 π₯ + π¦ = 50000 75π₯ + 120π¦ = 5190000 −75π₯ − 75π¦ = −3750000 75π₯ + 120π¦ = 5190000 45π¦ = 1440000 π¦ = 32000 π₯ + 32000 = 50000 → π₯ = 18000 $18,000 in real estate, $32,000 in small business 14) By using a graphing calculator, find the increasing interval(s) 1 for the function f x = π₯ 3 − 4π₯ 3 14) By using a graphing calculator, find the increasing interval(s) 1 for the function f x = π₯ 3 − 4π₯ 3 Choose windows large enough to allow you to see the turning points: π₯: −10, 10 π¦: −10, 10 −∞, −2 , (2, ∞) 15) Determine if the given function is even, odd, or neither. π₯2 + 2 π π₯ = 4 π₯ +1 15) Determine if the given function is even, odd, or neither. π₯2 + 2 π π₯ = 4 π₯ +1 (−π₯)2 +2 π −π₯ = (−π₯)4 +1 π₯2 + 2 = 4 π₯ +1 πΈπ£ππ π π₯+β −π π₯ 16) Determine the difference quotient, β 2 π π₯ = 3π₯ − 2π₯ + 5 , for the function π π₯+β −π π₯ 16) Determine the difference quotient, β 2 π π₯ = 3π₯ − 2π₯ + 5 , for the function π π₯+β −π π₯ β 2 3 π₯ + β − 2 π₯ + β + 5 − (3π₯ 2 − 2π₯ + 5) = β 3 π₯ 2 + 2π₯β + β2 − 2π₯ − 2β + 5 − 3π₯ 2 + 2π₯ − 5 = β 3π₯ 2 + 6π₯β + 3β2 − 2π₯ − 2β + 5 − 3π₯ 2 + 2π₯ − 5 = β 6π₯β + 3β2 − 2β = β = 6π₯ + 3β − 2 17) Suppose a state’s income tax code states that the tax liability T on x dollars of taxable income as follows T π₯ = 0.04π₯ 800 + 0.06π₯ ππ 0 ≤ π₯ < 20,000 ππ π₯ ≥ 20,000 Use π to find the tax liability on each taxable income: $12,000, $20,000, $50,000 17) Suppose a state’s income tax code states that the tax liability T on x dollars of taxable income as follows T π₯ = 0.04π₯ 800 + 0.06π₯ ππ 0 ≤ π₯ < 20,000 ππ π₯ ≥ 20,000 Use π to find the tax liability on each taxable income: $12,000, $20,000, $50,000 π 12000 = 0.04(12000) = $480 π 20000 = 800 + 0.06 20000 = $2000 π 50000 = 800 + 0.06 50000 = $3800 $480, $2000, $3800 18) Write an equation for the function whose graph fits the following description. The graph of π π₯ = π₯ 2 is shifted five units right, stretched vertically by a factor of three, reflected in the y-axis and shifted four units up. 18) Write an equation for the function whose graph fits the following description. The graph of π π₯ = π₯ 2 is shifted five units right, stretched vertically by a factor of three, reflected in the y-axis and shifted four units up. π βπππ‘ππ πππ£π π’πππ‘π πππβπ‘: π¦ = π₯ − 5 2 π π‘πππ‘πβππ π£πππ‘ππππππ¦ ππ¦ π ππππ‘ππ ππ π‘βπππ: π¦ = 3 π₯ − 5 πππππππ‘ππ ππ π‘βπ π¦ − ππ₯ππ : π¦ = 3 −π₯ − 5 π βπππ‘ππ 4 π’πππ‘π π’π: π¦ = 3 −π₯ − 5 π¦ = 3 −π₯ − 5 2 +4 2 +4 2 2 19) Find the vertex of the parabola π¦ = −2π₯ 2 + 10π₯ − 3. 19) Find the vertex of the parabola π¦ = −2π₯ 2 + 10π₯ − 3. π = −2, π = 10, π = −3 π 10 10 5 − =− =− = (ππ 2.5) 2π 2 β −2 −4 2 5 5 π = −2 2 2 =− 2 + 10 5 −3 2 25 + 25 − 3 2 19 = (ππ 9.5) 2 OR graph the function and find the maximum. 5 19 , ππ (2.5, 9.5) 2 2 20a) Find the vertical asymptotes, if any, of (π₯ − 3)(2π₯ − 2) π π₯ = (π₯ − 3)(π₯ + 4) 20a) Find the vertical asymptotes, if any, of (π₯ − 3)(2π₯ − 2) π π₯ = (π₯ − 3)(π₯ + 4) ππππ¦ π€βπππ π₯ + 4 = 0 π₯ = −4 (πππ‘ " − 4" → π¦ππ’ ππ’π π‘ π€πππ‘π π₯ = −4 π ππππ ππ‘ ππ π ππππ) 20b) Find the horizontal asymptotes, if any, of 2π₯ − 3 β π₯ = 3π₯ + 5 20b) Find the horizontal asymptotes, if any, of 2π₯ − 3 β π₯ = 3π₯ + 5 πππππππ πππ πππ’ππ 2 β¨π¦= 3 2 2 (πππ‘ → π¦ππ’ ππ’π π‘ π€πππ‘π π¦ = 3 3 π ππππ ππ‘ ππ π ππππ) 21) The safe load that a rectangular beam can support varies jointly as the width and square of the depth of the beam and inversely as its length. A beam 4 inches wide, 6 inches deep, and 25 feet long can support a safe load of 576 pounds. Find the safe load for a beam that is of the same material but is 6 inches wide, 10 inches deep, and 20 feet long. 21) The safe load that a rectangular beam can support varies jointly as the width and square of the depth of the beam and inversely as its length. A beam 4 inches wide, 6 inches deep, and 25 feet long can support a safe load of 576 pounds. Find the safe load for a beam that is of the same material but is 6 inches wide, 10 inches deep, and 20 feet long. ππ€π 2 π= π π(4)(62 ) 576 = 25 14400 = 144π π = 100 100π€π 2 100(6)(102 ) π= β¨π= = 3000 π 20 3000 πππ’πππ 22a) Solve. 3π₯ = 9 π₯−1 ⋅ 27 1−3π₯ 22a) Solve. 3π₯ = 9 π₯−1 ⋅ 27 π₯−1 1−3π₯ 1−3π₯ 3π₯ = 3 2 ⋅ 3 3 3π₯ = 32π₯−2 · 33−9π₯ 3π₯ = 31−7π₯ π₯ = 1 − 7π₯ 8π₯ = 1 1 π₯= 8 1 { } 8 22b) Solve. Please round your answer to three decimal places. 3π₯ −2 = 43−5π₯ 22b) Solve. Please round your answer to three decimal places. 3π₯ −2 = 43−5π₯ ln 3π₯−2 = ln 43−5π₯ π₯ − 2 ln 3 = 3 − 5π₯ ln 4 π₯ ln 3 − 2 ln 3 = 3 ln 4 − 5π₯ ln 4 π₯ ln 3 + 5π₯ ln 4 = 2 ln 3 + 3 ln 4 π₯(ln 3 + 5 ln 4) = 2 ln 3 + 3 ln 4 2 ln 3 + 3 ln 4 π₯= ≈ 0.792 ln 3 + 5 ln 4 {0.792} 23a) How much does Amy have to invest at 8% compounded continuously to have a balance of $19584.76 after ten years? 23a) How much does Amy have to invest at 8% compounded continuously to have a balance of $19584.76 after ten years? π΄ = ππ ππ‘ 19584.76 = ππ 0.08∗10 19584.76 = ππ 0.8 19584.76 π= ≈ 8800 0.8 π $8,800 23b) How long will it take an investment of $500 to double if the money is invested at 6% compounded monthly ? 23b) How long will it take an investment of $500 to double if the money is invested at 6% compounded monthly ? π π΄=π 1+ π ππ‘ 0.06 1000 = 500 1 + 12 12π‘ 2 = 1.00512π‘ ln 2 = ln 1.00512π‘ ln 2 = 12π‘ ln 1.005 ln 2 π‘= ≈ 11.58 12 ln 1.005 11.58 π¦ππππ 24a) Write the expression in expanded form. π¦+3 2 lπ[ ] 3 π₯ −2 24a) Write the expression in expanded form. π¦+3 2 lπ[ ] 3 π₯ −2 = ln π¦ + 3 2 − ln π₯ − 2 3 = 2 ln(π¦ + 3) − 3 ln(π₯ − 2) 24b) Write the expression in condensed form. 1 2 ln π₯ − ln(π₯ 2 + 1) 2 24b) Write the expression in condensed form. 1 2 ln π₯ − ln(π₯ 2 + 1) 2 1 2 ln π₯ − ln(π₯ 2 + 1) 2 = ln π₯ 2 − ln π₯ 2 + 1 1 2 = ln π₯ 2 − ln π₯ 2 + 1 = ln π₯2 π₯2 + 1 25a) Solve. log 2 5π₯ + 1 = −2 25a) Solve. log 2 5π₯ + 1 = −2 2−2 = 5π₯ + 1 1 = 5π₯ + 1 22 1 = 5π₯ + 1 4 1 5π₯ + 1 = 4 5π₯ = 1 −1 4 5π₯ = − π₯=− 3 4 3 20 1 π πππππππ − π‘βπππ πππ ππ ππππππ ππ π π’ππ π ππππ 5π₯ + 1 π€ππ π ππ‘ π‘π . 4 1 π₯ π€πππ πππ’π π 5π₯ + 1 π‘π ππ π€βππβ ππ πππππ‘ππ π‘βππ π§πππ. 4 {− 3 } 20 25b) Solve. log π₯ + 8 + log π₯ − 1 = 1 25b) Solve. log π₯ + 8 + log π₯ − 1 = 1 log π₯ + 8 + log(π₯ − 1) = 1 log π₯ + 8 π₯ − 1 = 1 101 = π₯ + 8 π₯ − 1 10 = π₯ 2 + 7π₯ − 8 π₯ 2 + 7π₯ − 18 = 0 π₯+9 π₯−2 = 0 π₯ = −9 π₯ = 2 π₯ ≠ −9 {2}