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Final Exam Schedule – Fall, 2014 11:00 ClassThursday, December 1110:30 – 12:30 2:00 ClassThursday, December 111:00 – 3:00 3:30 ClassTuesday, December 93:30 – 5:30

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66 –19 2x 2 + 10x – 7 Warm-up

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p(q(x)) = (x+4) 2 – 3x – 12 = x 2 + 8x + 16 – 3x – 12 = x 2 + 5x + 4 10. The answer to question 9 can be written as x 2 + 5x + 4. If that is not the answer you obtained, identify how this new answer was obtained. 5.4

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Last class, we did a power regression using years since 1900 and the equation of The power model was YearU.S. Population in millions 1950152.3 1960180.7 1970205.1 1980227.7 1990249.9 The number of automobiles A in the United States, in millions, is related to U.S. population P, in millions by the linear function A(P) =.59P – 36.7 Use the composite function to verify the number of automobiles in the United States in the year 2000. P(t) = 5.802(t.838 ) The table at right shows the population in the United States for the years indicated. Can we express the number of automobiles A in the United States as a function of years t since 1900? A(P(t)) =.59(5.802(t.838 )) – 36.7 or A = 3.423t.838 – 36.7 Approximately 125.6 million Based on the models, how many automobiles were in the U.S. in the year 2000? A = 3.423(100).838 – 36.7 125.6

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Bald Eagles were once very common throughout most of the United States. Their population numbers have been estimated at 300,000 to 500,000 birds in the early 1700s. Their population fell to threatened levels in the continental U.S. of less than 10,000 nesting pairs by the 1950s. Bald eagles were officially declared an endangered species in 1967 in all areas of the United States. Below is a table indicating the estimated number of breeding pairs of bald eagles in the lower 48 states in various years. Year Number of Breeding Pairs of Bald Eagles 19409551 19524750 19563210 19811188 19861875 19923749 20006471 Make a scatter plot of the data using years since 1940.

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Year Number of Breeding Pairs of Bald Eagles 19409551 19524750 19563210 19811188 19861875 19923749 20006471

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Standard Form: y = ax 2 + bx + c Quadratic Functions - Parabolas y =.5x 2 – 2x – 3 y = – x 2 + 2x + 4

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Standard Form: y = ax 2 + bx + c Quadratic Functions - Parabolas y =.5x 2 – 2x – 3 If a > 0, the parabola opens upward (vertex is a minimum). If a < 0, the parabola opens downward (vertex is a maximum). y = – x 2 + 2x + 4

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Standard Form: y = ax 2 + bx + c Quadratic Functions - Parabolas y =.5x 2 – 2x – 3 If a > 0, the parabola opens upward (vertex is a minimum). If a < 0, the parabola opens downward (vertex is a maximum). The larger the value of a is, the narrower the parabola. y = – x 2 + 2x + 4

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Standard Form: y = ax 2 + bx + c Quadratic Functions - Parabolas y =.5x 2 – 2x – 3 If a > 0, the parabola opens upward (vertex is a minimum). If a < 0, the parabola opens downward (vertex is a maximum). The larger the value of a is, the narrower the parabola. The y-intercept is at the point (0, c). y = – x 2 + 2x + 4

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Standard Form: y = ax 2 + bx + c Quadratic Functions - Parabolas y =.5x 2 – 2x – 3 If a > 0, the parabola opens upward (vertex is a minimum). If a < 0, the parabola opens downward (vertex is a maximum). The larger the value of a is, the narrower the parabola. The y-intercept is at the point (0, c). The zeros are the values of x at which the graph crosses the x-axis. y = – x 2 + 2x + 4

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Year Number of Breeding Pairs of Bald Eagles 19409551 19524750 19563210 19811188 19861875 19923749 20006471 2.Find the equation of a model relating the number of breeding pairs B to years t since 1940. Round coefficients to the nearest 100 th. 3. Based on the model, what was the number of breeding pairs of bald eagles in 1945? 4. Based on the model, in what year was the number of breeding pairs of bald eagles the lowest? What was the lowest number of breeding pairs in that year? 5. Based on the model, in what year did the number of breeding pairs of bald eagles again reach the 1940 level? 6. Verify your answer to question 5 algebraically.

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Year Number of Breeding Pairs of Bald Eagles 19409551 19524750 19563210 19811188 19861875 19923749 20006471 B = 7.94t 2 – 530.63t + 9695.69 where t = years since 1940, B = # of breeding pairs 2.Find the equation of a model relating the number of breeding pairs B to years t since 1940. Round coefficients to the nearest 100 th. 3. Based on the model, what was the number of breeding pairs of bald eagles in 1945? 4. Based on the model, in what year was the number of breeding pairs of bald eagles the lowest? What was the lowest number of breeding pairs in that year? 5. Based on the model, in what year did the number of breeding pairs of bald eagles again reach the 1940 level? 6. Verify your answer to question 5 algebraically.

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Year Number of Breeding Pairs of Bald Eagles 19409551 19524750 19563210 19811188 19861875 19923749 20006471 7,241 breeding pairs 2.Find the equation of a model relating the number of breeding pairs B to years t since 1940. Round coefficients to the nearest 100 th. 3. Based on the model, what was the number of breeding pairs of bald eagles in 1945? 4. Based on the model, in what year was the number of breeding pairs of bald eagles the lowest? What was the lowest number of breeding pairs in that year? 5. Based on the model, in what year did the number of breeding pairs of bald eagles again reach the 1940 level? 6. Verify your answer to question 5 algebraically. y = 794x 2 – 530.63x + 9695.69 where x = years since 1940, y = # of breeding pairs B = 7.94t 2 – 530.63t + 9695.69 where t = years since 1940, B = # of breeding pairs

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Year Number of Breeding Pairs of Bald Eagles 19409551 19524750 19563210 19811188 19861875 19923749 20006471 1973 had the lowest number of breeding pairs at about 830 7,241 breeding pairs 2.Find the equation of a model relating the number of breeding pairs B to years t since 1940. Round coefficients to the nearest 100 th. 3. Based on the model, what was the number of breeding pairs of bald eagles in 1945? 4. Based on the model, in what year was the number of breeding pairs of bald eagles the lowest? What was the lowest number of breeding pairs in that year? 5. Based on the model, in what year did the number of breeding pairs of bald eagles again reach the 1940 level? 6. Verify your answer to question 5 algebraically. y = 7.94x 2 – 530.63x + 9695.69 where x = years since 1940, y = # of breeding pairs B = 7.94t 2 – 530.63t + 9695.69 where t = years since 1940, B = # of breeding pairs

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Year Number of Breeding Pairs of Bald Eagles 19409551 19524750 19563210 19811188 19861875 19923749 20006471 1973 had the lowest number of breeding pairs at about 830 7,241 breeding pairs 2.Find the equation of a model relating the number of breeding pairs B to years t since 1940. Round coefficients to the nearest 100 th. 3. Based on the model, what was the number of breeding pairs of bald eagles in 1945? 4. Based on the model, in what year was the number of breeding pairs of bald eagles the lowest? What was the lowest number of breeding pairs in that year? 5. Based on the model, in what year did the number of breeding pairs of bald eagles again reach the 1940 level? 6. Verify your answer to question 5 algebraically. 2006 or 2007 9551 = 7.94t 2 – 530.63x + 9695.69

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To solve for x in a quadratic equation 1. Set one side equal to zero 2. Use the Quadratic Formula: The solutions to the equation ax 2 + bx + c = 0 are

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Year Number of Breeding Pairs of Bald Eagles 19409551 19524750 19563210 19811188 19861875 19923749 20006471 1973 had the lowest number of breeding pairs at about 830 7,241 breeding pairs 2.Find the equation of a model relating the number of breeding pairs B to years t since 1940. Round coefficients to the nearest 100 th. 3. Based on the model, what was the number of breeding pairs of bald eagles in 1945? 4. Based on the model, in what year was the number of breeding pairs of bald eagles the lowest? What was the lowest number of breeding pairs in that year? 5. Based on the model, in what year will the number of breeding pairs of bald eagles again reach the 1940 level? 6. Verify your answer to question 5 algebraically. 2006 or 2007 9551 = 7.94t 2 – 530.63t + 9695.69

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0 = 7.94t 2 – 530.63t + 144.69 -9551 a = 7.94 t = 66.55,.27 2006 1940 b = –530.63 c = 144.69

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Research by the automobile industry has shown that driving speeds affect gas mileage. The table at the right shows the average gas mileage G (in miles per gallon) for cars traveling at different speeds, S (in miles per hour). SpeedMiles per gallon 4022.0 4523.8 5025.5 6125.3 6623.9 7022.6 Make a scatterplot of gas mileage vs. speed.

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Without Vars With Vars

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Research by the automobile industry has shown that driving speeds affect gas mileage. The table at the right shows the average gas mileage G (in miles per gallon) for cars traveling at different speeds, S (in miles per hour). 1. Find a quadratic model for the data. Round coefficients to the nearest thousandth (three decimal places). 2.Based on the model, how many miles per gallon would you expect a car traveling at an average speed of 75 miles per hour to get? 3.To the nearest tenth, what speed gives the maximum gas mileage? 4. At what speed (nearest tenth) will a car get exactly 23 miles per gallon? Give an algebraic solution. G = -.016S 2 +1.750S – 22.859 SpeedMiles per gallon 4022.0 4523.8 5025.5 6125.3 6623.9 7022.6

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Research by the automobile industry has shown that driving speeds affect gas mileage. The table at the right shows the average gas mileage G (in miles per gallon) for cars traveling at different speeds, S (in miles per hour). 1. Find a quadratic model for the data. Round coefficients to the nearest thousandth (three decimal places). 2.Based on the model, how many miles per gallon would you expect a car traveling at an average speed of 75 miles per hour to get? 3.To the nearest tenth, what speed gives the maximum gas mileage? 4. At what speed (nearest tenth) will a car get exactly 23 miles per gallon? Give an algebraic solution. G = -.016S 2 +1.750S – 22.859 18.4 miles per gallon (w/o Vars) 19.8 miles per gallon (with Vars) SpeedMiles per gallon 4022.0 4523.8 5025.5 6125.3 6623.9 7022.6

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Research by the automobile industry has shown that driving speeds affect gas mileage. The table at the right shows the average gas mileage G (in miles per gallon) for cars traveling at different speeds, S (in miles per hour). 1. Find a quadratic model for the data. Round coefficients to the nearest thousandth (three decimal places). 2.Based on the model, how many miles per gallon would you expect a car traveling at an average speed of 75 miles per hour to get? 3.To the nearest tenth, what speed gives the maximum gas mileage? 4. At what speed (nearest tenth) will a car get exactly 23 miles per gallon? Give an algebraic solution. G = -.016S 2 +1.750S – 22.859 18.4 miles per gallon (w/o Vars) 19.8 miles per gallon (with Vars) 54.7 miles per hour (w/o Vars) 55.6 miles per hour (with Vars) SpeedMiles per gallon 4022.0 4523.8 5025.5 6125.3 6623.9 7022.6

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0 = -.016S 2 +1.750S – 45.859 -23 a = -.016 x = 43.5, 65.875 b = 1.75 c = -45.859 23 = -.016S 2 +1.750S – 22.859G

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Research by the automobile industry has shown that driving speeds affect gas mileage. The table at the right shows the average gas mileage G (in miles per gallon) for cars traveling at different speeds, S (in miles per hour). 1. Find a quadratic model for the data. Round coefficients to the nearest thousandth (three decimal places). 2.Based on the model, how many miles per gallon would you expect a car traveling at an average speed of 75 miles per hour to get? 3.To the nearest tenth, what speed gives the maximum gas mileage? 4. At what speed (nearest tenth) will a car get exactly 23 miles per gallon? Give an algebraic solution. G = -.016S 2 +1.750S – 22.859 18.4 miles per gallon (w/o Vars) 19.8 miles per gallon (with Vars) 54.7 miles per hour (w/o Vars) 55.6 miles per hour (with Vars) SpeedMiles per gallon 4022.0 4523.8 5025.5 6125.3 6623.9 7022.6 43.5 mph and 65.9 mph

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Homework: Complete Parts II and III Section 5.4 (Composite Functions) AND Section 5.5 – Quadratic Functions

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MAT 150 – Algebra Class #11 Topics:

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