This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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You're on a roll — To plot these exponential functions, we'll identify key features like the horizontal asymptote, y-intercept, and a few points. a) f(x) = -2^-x + 4 This function can be rewritten as f(x) = -((1)/(2))^x + 4. Step 1: Identify the horizontal asymptote. The term +4 indicates a vertical shift upwards by 4 units. The horizontal asymptote is at y = 4. Step 2: Find the y-intercept. Set x=0: f(0) = -2^-0 + 4 = -1 + 4 = 3 The y-intercept is (0, 3). Step 3: Calculate additional points for plotting. | x | f(x) = -2^-x + 4 | Point | | :-- | :-------------------- | :---- | | -2 | -2^-(-2) + 4 = -2^2 + 4 = -4 + 4 = 0 | (-2, 0) | | -1 | -2^-(-1) + 4 = -2^1 + 4 = -2 + 4 = 2 | (-1, 2) | | 0 | -2^0 + 4 = -1 + 4 = 3 | (0, 3) | | 1 | -2^-1 + 4 = -(1)/(2) + 4 = 3.5 | (1, 3.5) | | 2 | -2^-2 + 4 = -(1)/(4) + 4 = 3.75 | (2, 3.75) | Step 4: State the domain and range. The domain is all real numbers: x (-, ). The range is all real numbers less than the horizontal asymptote: y (-, 4). b) f(x) = -2(2)^x + 2 Step 1: Identify the horizontal asymptote. The term +2 indicates a vertical shift upwards by 2 units. The horizontal asymptote is at y = 2. Step 2: Find the y-intercept. Set x=0: f(0) = -2(2)^0 + 2 = -2(1) + 2 = -2 + 2 = 0 The y-intercept is (0, 0). Step 3: Calculate additional points for plotting. | x | f(x) = -2(2)^x + 2 | Point | | :-- | :-------------------- | :---- | | -2 | -2(2)^-2 + 2 = -2((1)/(4)) + 2 = -(1)/(2) + 2 = 1.5 | (-2, 1.5) | | -1 | -2(2)^-1 + 2 = -2((1)/(2)) + 2 = -1 + 2 = 1 | (-1, 1) | | 0 | -2(2)^0 + 2 = -2(1) + 2 = 0 | (0, 0) | | 1 | -2(2)^1 + 2 = -4 + 2 = -2 | (1, -2) | | 2 | -2(2)^2 + 2 = -2(4) + 2 = -8 + 2 = -6 | (2, -6) | Step 4: State the domain and range. The domain is all real numbers: x (-, ). The range is all real numbers less than the horizontal asymptote: y (-, 2). c) f(x) = 3((1)/(2))^x + 2 Step 1: Identify the horizontal asymptote. The term +2 indicates a vertical shift upwards by 2 units. The horizontal asymptote is at y = 2. Step 2: Find the y-intercept. Set x=0: f(0) = 3((1)/(2))^0 + 2 = 3(1) + 2 = 3 + 2 = 5 The y-intercept is (0, 5). Step 3: Calculate additional points for plotting. | x | f(x) = 3((1)/(2))^x + 2 | Point | | :-- | :-------------------- | :---- | | -2 | 3((1)/(2))^-2 + 2 = 3(4) + 2 = 12 + 2 = 14 | (-2, 14) | | -1 | 3((1)/(2))^-1 + 2 = 3(2) + 2 = 6 + 2 = 8 | (-1, 8) | | 0 | 3((1)/(2))^0 + 2 = 3(1) + 2 = 5 | (0, 5) | | 1 | 3((1)/(2))^1 + 2 = 3((1)/(2)) + 2 = 1.5 + 2 = 3.5 | (1, 3.5) | | 2 | 3((1)/(2))^2 + 2 = 3((1)/(4)) + 2 = 0.75 + 2 = 2.75 | (2, 2.75) | Step 4: State the domain and range. The domain is all real numbers: x (-, ). The range is all real numbers greater than the horizontal asymptote: y (2, ). What's next?