for 9 -17, sketch the graph of each exponential function by doing the following: sketch the asymptote, label at least two distinct coordinate points on each graph, and write the domain and range of each function. 9. $y = 2(4)^x$ 10. $y = -(3)^x$ 11. $y = -2(3)^x$ domain: range: domain: range: domain: range: 12. $y = 4(2)^x - 3$ 13. $y = -3(2)^{x - 1}$ 14. $y = 3(3)^{x + 2} - 4$ domain: range: domain: range: domain: range:

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### Problem 9: $ y = 2(4)^x $ # Explanation: ## Step 1: Find the Asymptote For an exponential function of the form $ y = ab^x + k $, the horizontal asymptote is $ y = k $. Here, $ k = 0 $, so the asymptote is $ y = 0 $ (the x - axis). ## Step 2: Find Coordinate Points - When $ x = 0 $: $ y=2(4)^0=2\times1 = 2 $, so the point is $ (0, 2) $. - When $ x = 1 $: $ y = 2(4)^1=2\times4=8 $, so the point is $ (1, 8) $. ## Step 3: Determine Domain and Range - Domain: Exponential functions of the form $ y = ab^x $ have a domain of all real numbers, so Domain: $ (-\infty, \infty) $ - Range: Since $ 4^x>0 $ for all real $ x $, then $ 2(4)^x>0 $. So Range: $ (0, \infty) $ # Answer: Asymptote: $ y = 0 $ Points: $ (0, 2) $, $ (1, 8) $ Domain: $ (-\infty, \infty) $ Range: $ (0, \infty) $ ### Problem 10: $ y=-(3)^x $ # Explanation: ## Step 1: Find the Asymptote For $ y = - 3^x $, the form is $ y=ab^x + k $ with $ k = 0 $, so the horizontal asymptote is $ y = 0 $ (the x - axis). ## Step 2: Find Coordinate Points - When $ x = 0 $: $ y=-(3)^0=- 1\times1=-1 $, so the point is $ (0, - 1) $ - When $ x = 1 $: $ y=-(3)^1=-3 $, so the point is $ (1, - 3) $ ## Step 3: Determine Domain and Range - Domain: For $ y=-3^x $, the domain is all real numbers, so Domain: $ (-\infty, \infty) $ - Range: Since $ 3^x>0 $ for all real $ x $, then $ - 3^x<0 $. So Range: $ (-\infty, 0) $ # Answer: Asymptote: $ y = 0 $ Points: $ (0, - 1) $, $ (1, - 3) $ Domain: $ (-\infty, \infty) $ Range: $ (-\infty, 0) $ ### Problem 11: $ y=-2(3)^x $ # Explanation: ## Step 1: Find the Asymptote For $ y=-2(3)^x $, the form is $ y = ab^x+k $ with $ k = 0 $, so the horizontal asymptote is $ y = 0 $ (the x - axis). ## Step 2: Find Coordinate Points - When $ x = 0 $: $ y=-2(3)^0=-2\times1=-2 $, so the point is $ (0, - 2) $ - When $ x = 1 $: $ y=-2(3)^1=-2\times3 = - 6 $, so the point is $ (1, - 6) $ ## Step 3: Determine Domain and Range - Domain: All real numbers, so Domain: $ (-\infty, \infty) $ - Range: Since $ 3^x>0 $, then $ - 2(3)^x<0 $. So Range: $ (-\infty, 0) $ # Answer: Asymptote: $ y = 0 $ Points: $ (0, - 2) $, $ (1, - 6) $ Domain: $ (-\infty, \infty) $ Range: $ (-\infty, 0) $ ### Problem 12: $ y = 4(2)^x-3 $ # Explanation: ## Step 1: Find the Asymptote For $ y = 4(2)^x-3 $, the horizontal asymptote is $ y=-3 $ (since the form is $ y = ab^x + k $ with $ k=-3 $) ## Step 2: Find Coordinate Points - When $ x = 0 $: $ y=4(2)^0-3=4\times1 - 3=1 $, so the point is $ (0, 1) $ - When $ x = 1 $: $ y=4(2)^1-3=4\times2-3 = 5 $, so the point is $ (1, 5) $ ## Step 3: Determine Domain and Range - Domain: All real numbers, so Domain: $ (-\infty, \infty) $ - Range: Since $ 2^x>0 $, then $ 4(2)^x>0 $, and $ 4(2)^x-3>-3 $. So Range: $ (-3, \infty) $ # Answer: Asymptote: $ y=-3 $ Points: $ (0, 1) $, $ (1, 5) $ Domain: $ (-\infty, \infty) $ Range: $ (-3, \infty) $ ### Problem 13: $ y=-3(2)^{x - 1} $ # Explanation: ## Step 1: Find the Asymptote For $ y=-3(2)^{x - 1} $, the form is $ y = ab^x + k $ with $ k = 0 $ (rewriting $ y=-3\times2^{-1}\times2^x=-\frac{3}{2}(2)^x $), so the horizontal asymptote is $ y = 0 $ ## Step 2: Find Coordinate Points - When $ x = 1 $: $ y=-3(2)^{1 - 1}=-3(2)^0=-3\times1=-3 $, so the point is $ (1, - 3) $ - When $ x = 2 $: $ y=-3(2)^{2 - 1}=-3(2)^1=-3\times2=-6 $, so the point is $ (2, - 6) $ ## Step 3: Determine Domain and Range - Domain: All real numbers, so Domain: $ (-\infty, \infty) $ - Range: Since $ 2^{x-1}>0 $ for all real $ x $, then $ - 3(2)^{x - 1}<0 $. So Range: $ (-\infty, 0) $ # Answer: Asymptote: $ y = 0 $ Points: $ (1, - 3) $, $ (2, - 6) $ Domain: $ (-\infty, \infty) $ Range: $ (-\infty, 0) $ ### Problem 14: $ y = 3(3)^{x + 2}-4 $ # Explanation: ## Step 1: Find the Asymptote For $ y = 3(3)^{x + 2}-4 $, we can rewrite it as $ y=3\times3^2\times3^x-4 = 27(3)^x-4 $. The form is $ y = ab^x + k $ with $ k=-4 $, so the horizontal asymptote is $ y=-4 $ ## Step 2: Find Coordinate Points - When $ x=-2 $: $ y=3(3)^{-2 + 2}-4=3(3)^0-4=3\times1-4=-1 $, so the point is $ (-2, - 1) $ - When $ x=-1 $: $ y=3(3)^{-1 + 2}-4=3(3)^1-4=9 - 4 = 5 $, so the point is $ (-1, 5) $ ## Step 3: Determine Domain and Range - Domain: All real numbers, so Domain: $ (-\infty, \infty) $ - Range: Since $ 3^{x + 2}>0 $ for all real $ x $, then $ 3(3)^{x + 2}>0 $, and $ 3(3)^{x + 2}-4>-4 $. So Range: $ (-4, \infty) $ # Answer: Asymptote: $ y=-4 $ Points: $ (-2, - 1) $, $ (-1, 5) $ Domain: $ (-\infty, \infty) $ Range: $ (-4, \infty) $

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Explanation:

Problem 9: \( y = 2(4)^x \)

Step 1: Find the Asymptote

Step 2: Find Coordinate Points

Step 3: Determine Domain and Range

Step 1: Find the Asymptote

Step 2: Find Coordinate Points

Step 3: Determine Domain and Range

Step 1: Find the Asymptote

Step 2: Find Coordinate Points

Step 3: Determine Domain and Range

Answer:

Problem 10: \( y=-(3)^x \)