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20:51:36 5.3.38. Explain how you used transformations to graph f(x) = 2^(x+2)
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RESPONSE --> first graph y - 2^x from there shift graph left 2 units
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20:53:25 The function y = 2 ^ ( x+2) is a transformation of the basic function y = 2^x, with x replaced by x - 2. So the graph moves 2 units in the x direction. The graph of y = 2^x is asymptotic to the negative x axis, increases at a rapidly increasing rate and passes through ( 0, 1 ) and (1, 2). The graph of y = 2^(x-2), being shifted in only the x direction, is also asymptotic to the negative x axis and passes through the points (0 + 2, 1) = (2, 1) and (1 + 2, 2) = (3, 2). The graph also increases at a rapidly increasing rate. All the points of the graph of y = 2^(x-2) lie 2 units to the right of points on the graph of y = 2^x.
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RESPONSE --> the book has 2^(x+2) which is why I had to shift 2 units to the left.
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20:55:12 5.3.42. Transformations to graph f(x) = 1 – 3 * 2^x
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RESPONSE --> first graph y=2^x then mulitply by -3, reflect about the x-axis, and shift up one unit.
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20:56:23 1 - 3 * 2^x is obtained from y = 2^x by first vertically stretching the graph by factor -3, then by shifting this graph 1 unit in the vertical direction. The graph of y = 2^x is asymptotic to the negative x axis, increases at a rapidly increasing rate and passes through ( 0, 1 ) and (1, 2). -3 * 2^x will approach zero as 2^x approaches zero, so the graph of y = -3 * 2^x will remain asymptotic to the negative x axis. The points (0, 1) and (1, 2) will be transformed to (0, -3* 1) = (0, -3) and (1, -3 * 2) = (1, -6). So the graph will decrease from its asymptote just below the x axis through the points (0, -3) and (1, -6), decreasing at a rapidly decreasing rate. To get the graph of 1 - 3 * 2^x the graph of -3 * 2^x will be vertically shifted 1 unit. This will raise the horizontal asymptote from the x axis (which is the line y = 0) to the line y = 1, and will also raise every other point by 1 unit. The points (0, -3) and (1, -6) will be transformed to (0, -3 + 1) = (0, -2) and (1, -6 + 5) = (1, -5), decreasing from its asymptote just below the line y = 1 through the points (0, -2) and (1, -5), decreasing at a rapidly decreasing rate.
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RESPONSE --> ok
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20:58:24 5.3.60 Solve (1/2)^(1-x) = 4.
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RESPONSE --> (1/2)^(1-x) = 2^2 (2^-1)^(1-x) = 2^2 2^(-1+x)=2^2 -1+x=2 x=3
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20:59:23 (1/2)^(-2) = 1 / (1/2)^2 = 1 / (1/4) = 4, so we know that the exponennt 1 - x must be -2. If 1 - x = -2, it follows that -x = -2 - 1 = -3 and x = 3.
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RESPONSE --> I solved mine differently than you did yours.
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21:01:20 5.3.78 A(n) = A0 e^(-.35 n), area of wound after n days. What is the area after 3 days and what is the area after 10 days? Init area is 100 mm^2.
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RESPONSE --> A(3) = 100e^(.35*3) = 35mm^2 A(10) = 100e^(-0.35*10) =3.02 mm^2
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21:02:15 If the initial area is 100 mm^2, then when n = 0 we have A(0) = A0 e^(-.35 * 0) = 100 mm^2. Since e^0 = 1 this tells us that A0 = 100 mm^2. So the function is A(n) = 100 mm^2 * e^(-.35 n). To get the area after 3 days we evaluate the function for n = 3, obtainind A(3) = 100 mm^2 * e^(-.35 * 3) = 100 mm^3 * .35 = 35 mm^2 approx.. After to days we find that the area is A(10) = 100 mm^2 * e^(-.35 * 10) 100 mm63 * .0302 = 3.02 mm^2 approx..
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RESPONSE --> right
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21:04:27 5.3.84. Poisson probability 4^x e^-4 / x!, probability that x people will arrive in the next minute. What is the probability that 5 will arrive in the next minute, and what is the probability that 8 will arrive in the next minute?
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RESPONSE --> for x=5: (4^5 e^-4)/8! = .15629 = 15.6% for x = 8 (4^8 e^-4)/8! = .02977 = 2.97%
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21:04:53 he probability that 5 will arrive in the next minute is P(5) = 4^5 * e^-4 / (5 !) = 1024 * .0183 / (5 * 4 * 3 * 2 * 1) = 1024 * .0183 / 120 = .152 approx..
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RESPONSE --> ok
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21:06:23 Te probability that 8 will arrive in the next minute is P(8) = 4^8 * e^-8 / (8 !) = .00055 approx..
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RESPONSE --> e does not have a variable exponent in this problem. the formula given to us is P(x) = (4^x*e^-4)/x!
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