Math141Sols2S11

Math141Sols2S11 - MIDTERM 2 SOLUTIONS (CHAPTERS 2 AND 3)...

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MIDTERM 2 SOLUTIONS (CHAPTERS 2 AND 3) MATH 141 – SPRING 2011 – KUNIYUKI 150 POINTS TOTAL: 56 FOR PART 1, AND 94 FOR PART 2 Show all work, simplify as appropriate, and use “good form and procedure” (as in class). Box in your final answers! No notes or books allowed. Unless otherwise specified, give exact answers. Write units where appropriate in your answers. PART 1: SCIENTIFIC CALCULATORS ALLOWED! (56 POINTS TOTAL) 1) Write the “Vertex Form” of the equation of the parabola in the usual xy -plane that has ± 1, 4 () as its vertex and that passes through the point 1, ± 16 . (7 points) Remember that Vertex Form is given by: y = ax ± h 2 + k . Here, the vertex h , k = ± . We now have: y = ±± 1 2 + 4 y = + 1 2 + 4 Now, apply the Basic Principle of Graphing and use the fact that x = y = ± 16 must satisfy the equation of the parabola: y = + 1 2 + 4 ± ² 16 = a 1 + 1 2 + 4 ² 16 = a 2 2 + 4 ² 16 = 4 a + 4 ² 20 = 4 a a = ² 5 ± y = ² 5 x + 1 2 + 4
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2) An astronaut kicks a ball over a flat region of a (very) distant moon. The height of the ball in feet is given by: ht () = ± 3 t 2 + 18 t + 2i f t ² 0 , where t is the amount of time in seconds since the ball was kicked. (The formula is relevant up until the time the ball hits the ground.) (19 points total) a) Use a formula we used in class to find how much time it takes (since the ball was kicked) for the ball to reach its maximum height. The graph of h versus t is a parabola opening downward. We want the t -coordinate corresponding to the vertex (i.e., the maximum point) of the parabola. t = ± b 2 a = ± 18 2 ± 3 = 3 seconds after the ball was kicked b) What is the maximum height achieved by the ball? The height of the ball 3 seconds after it was kicked is given by: h 3 = ± 33 2 + 18 3 + 2 = ± 27 + 54 + 2 = 29 feet c) What was the height of the ball at the time it was kicked? h 0 = 2 feet
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d) How much time does it take (since the ball was kicked) for the ball to hit the ground? Round off your answer to three significant digits. Solve ht () = 0 for t > 0 ± 3 t 2 + 18 t + 2 = 0 for t > 0 We want the positive real zero of ± 3 t 2 + 18 t + 2 . The discriminant of ± 3 t 2 + 18 t + 2 a = ± 3, b = 18, c = 2 is: b 2 ± 4 ac = 18 2 ± 4 ± 3 2 = 324 + 24 = 348 This is not a “nice” square, so traditional factoring will not work. Let’s use the Quadratic Formula (QF). t = ± b ± 2 ± 4 ac 2 a = ± 18 ± 348 2 ± 3 ² We found the discriminant already. (You can use your calculator to find an approximate answer now.) = ± 18 ± 7 ± 6 = ± 18 ± 6 ± 7 ± 6 = 3 ± ± 87 3 ³ ´ µ · ¸ = 3 ± 87 3 or 9 ± 87 3 Observe: 3 ± 87 3 ²± 0.109 < 0 , so we discard this zero. Our answer is given by: 3 + 87 3 or 9 + 87 3 ± 6.11 Answer: About 6.11 seconds.
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Note : The graph of h against t is below. Observe that the h -intercept is 0, 2 () , and the t -intercept is about 6.11, 0 . 3) Consider fx = 3 x 3 ± 10 x 2 + 7 x ± 12 . Hint: One of the zeros is 3. (16 points) a) Write the two other complex zeros of f in simplest, standard form.
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Math141Sols2S11 - MIDTERM 2 SOLUTIONS (CHAPTERS 2 AND 3)...

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