Which graph represents the function [tex]y=5 \cdot\left(\frac{1}{3}\right)^x[/tex]?
Answer (2)
The function y = 5 ⋅ ( 3 1 ) x is a decreasing exponential function with the y-intercept at (0, 5) and a horizontal asymptote at y=0. As x increases, the value of y approaches 0. In contrast, as x goes negative, y increases significantly.
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The function is an exponential decay function with a base of 3 1 and a vertical stretch of 5.
The y-intercept is found by setting x = 0 , which gives y = 5 .
As x approaches infinity, y approaches 0, indicating a horizontal asymptote at y = 0 .
The graph is decreasing and passes through the point (0, 5), so the answer is y = 5" , ( 1/3 ) x .
Explanation
Understanding the Function The problem asks us to identify the graph of the function y = 5" , ( 1/3 ) x . This is an exponential function with a base of 3 1 , which is between 0 and 1. This tells us that the function is decreasing. Also, the coefficient 5 stretches the graph vertically.
Finding the y-intercept To sketch the graph, we can find some key points. First, let's find the y-intercept by setting x = 0 : y = 5" , ( 1/3 ) 0 = 5" , 1 = 5 So, the y-intercept is at the point (0, 5).
Asymptotic Behavior as x approaches infinity Next, let's consider the behavior of the function as x approaches infinity. As x → ∞ , ( 3 1 ) x → 0 , so y → 5 ⋅ 0 = 0 . This means the x-axis (y=0) is a horizontal asymptote.
Asymptotic Behavior as x approaches negative infinity Now, let's consider the behavior of the function as x approaches negative infinity. As x → − ∞ , ( 3 1 ) x → ∞ , so y → ∞ . This means the function increases without bound as x becomes more negative.
Summary of Graph Characteristics Based on this analysis, the graph should pass through (0, 5), be decreasing, have a horizontal asymptote at y=0, and increase without bound as x approaches negative infinity.
Calculating Additional Points To further confirm the shape, we can calculate a few more points:
When x = 1 , y = 5" , ( 1/3 ) 1 = 3 5 ≈ 1.67 When x = 2 , y = 5" , ( 1/3 ) 2 = 5" , 9 1 = 9 5 ≈ 0.56 When x = − 1 , y = 5" , ( 1/3 ) − 1 = 5" , 3 = 15 When x = − 2 , y = 5" , ( 1/3 ) − 2 = 5" , 9 = 45
These points confirm the decreasing nature and asymptotic behavior of the function.
Conclusion Therefore, the graph that represents the function y = 5" , ( 1/3 ) x is a decreasing exponential function that passes through (0, 5) and has a horizontal asymptote at y=0.
Examples
Exponential decay functions, like the one in this problem, are used to model various real-world phenomena. For example, they can describe the decay of radioactive substances, the cooling of an object, or the depreciation of an asset over time. Understanding the properties of exponential decay helps in predicting and analyzing these processes. For instance, if you invest in a car, its value decreases over time, and this depreciation can be modeled using an exponential decay function. Similarly, in medicine, the concentration of a drug in the bloodstream decreases exponentially over time.
