Graphing Exponential Functions: F(x) = 60(1/3)^x Explained

When we dive into the world of functions, especially exponential functions, understanding their graphical representation is key. Today, we're going to dissect the function $f(x)=60\left(\frac{1}{3}\right)^x$ and figure out what its graph looks like. Forget those boring textbook descriptions; we're going to break it down in a way that makes sense, so you can confidently interpret and visualize these kinds of graphs. Let's start by looking at the initial value and how the function behaves over time. The initial value, often denoted as $f(0)$ or $y_0$, is simply the value of the function when $x=0$. In our case, $f(0) = 60\left(\frac{1}{3}\right)^0$. Remember, any non-zero number raised to the power of zero is 1. So, $f(0) = 60 \times 1 = 60$. This means the graph starts at the point (0, 60). This is a crucial piece of information because it tells us where our graph intercepts the y-axis. It’s the starting point of our journey on the coordinate plane. Now, let’s talk about the "successive term" part. In exponential functions, we don't add or subtract a constant value; instead, we multiply by a constant factor, called the base. In $f(x)=60\left(\frac{1}{3}\right)^x$, the base is $\frac{1}{3}$. This base dictates how the function grows or decays. Since our base is $\frac{1}{3}$, which is less than 1 but greater than 0, our function will decay exponentially. This means as $x$ increases, the value of $f(x)$ gets smaller and smaller, approaching zero but never quite reaching it. Conversely, as $x$ decreases (becomes more negative), the value of $f(x)$ will increase rapidly. It's like a snowball rolling down a hill, but in reverse! The initial value of 60 is multiplied by $\frac{1}{3}$ for each unit increase in $x$. So, at $x=1$, $f(1) = 60\left(\frac{1}{3}\right)^1 = 20$. At $x=2$, $f(2) = 60\left(\frac{1}{3}\right)^2 = 60\left(\frac{1}{9}\right) = \frac{60}{9} = \frac{20}{3} \approx 6.67$. See how the values are decreasing? This decay is the hallmark of an exponential function with a base between 0 and 1. Let's consider the options provided, though they seem to have a misunderstanding of the initial value and the operation. Option A suggests an initial value of 20 and subtraction, which is incorrect. Option B also mentions an initial value of 20, which is wrong; our initial value is 60. The description of how successive terms are determined is also critical. It's not subtraction; it's multiplication by the base. The phrase "each successive term is determined by subtracting $\frac{1}{3}$" describes an arithmetic sequence, not an exponential function. For an exponential function like ours, $f(x)=60\left(\frac{1}{3}\right)^x$, the value of the function at $x+1$ is obtained by multiplying the value at $x$ by the base. That is, $f(x+1) = f(x) \times \frac{1}{3}$. This multiplicative relationship is what defines exponential behavior. So, to correctly describe the graph of $f(x)=60\left(\frac{1}{3}\right)^x$: it has an initial value of 60 (where it crosses the y-axis at $x=0$), and as $x$ increases, the function values decrease exponentially because the base $\frac{1}{3}$ is between 0 and 1. The graph will approach the x-axis (y=0) as $x$ goes to infinity, and it will rise steeply as $x$ becomes more negative. Understanding these core components – the initial value (y-intercept) and the base – allows us to accurately sketch and interpret any exponential function's graph.

Decoding the Components: Initial Value and Base

Let's dive a bit deeper into the two crucial components of our exponential function, $f(x)=60\left(\frac{1}{3}\right)^x$: the initial value and the base. Grasping these elements is like having a secret decoder ring for understanding the entire graph. First, the initial value. In the general form of an exponential function, $f(x) = a imes b^x$, the coefficient '$a$' represents the initial value. This is the value of the function when $x=0$. For our specific function, $f(x)=60\left(\frac{1}{3}\right)^x$, the initial value '$a$' is 60. This means that when $x=0$, $f(0) = 60 \times \left(\frac{1}{3}\right)^0 = 60 \times 1 = 60$. So, the graph of this function will always pass through the point (0, 60) on the y-axis. This is your y-intercept, the starting point from which all other values are derived through multiplication. It’s fundamental because it anchors the graph. Now, let's shift our focus to the base, which is '$b$' in the general form. In our function, the base is $\frac{1}{3}$. The base is the number that is repeatedly multiplied by itself, as indicated by the exponent '$x$'. The value of the base tells us about the function's behavior: whether it grows or decays, and how quickly. Since our base, $\frac{1}{3}$, is less than 1 (but greater than 0), this function exhibits exponential decay. Think of it this way: each time $x$ increases by 1, the function's value is multiplied by $\frac{1}{3}$. So, if we start at $x=0$ with $f(0)=60$, then at $x=1$, $f(1) = 60 \times \frac{1}{3} = 20$. At $x=2$, $f(2) = 20 \times \frac{1}{3} = \frac{20}{3}$, and so on. The values are getting progressively smaller. The graph will descend from left to right, getting closer and closer to the x-axis. This downward trend is characteristic of decay. If the base were greater than 1 (e.g., 2 or 3), the function would exhibit exponential growth, meaning the graph would rise from left to right. If the base were exactly 1, the function would be constant ($f(x) = a imes 1^x = a$), resulting in a horizontal line. If the base were negative, the graph would oscillate, which is not typically what we mean by standard exponential functions in this context. Therefore, the combination of an initial value of 60 and a base of $\frac{1}{3}$ tells us that the graph begins at (0, 60) and slopes downwards, approaching the x-axis as $x$ increases. It's a visual story of starting at a certain point and diminishing over time. The options provided in the prompt incorrectly identify the initial value and confuse the nature of exponential change with arithmetic change. An initial value of 20 is incorrect, as we've clearly calculated it to be 60. Furthermore, describing the change as "subtracting $\frac{1}{3}$" is fundamentally wrong; it should be multiplying by $\frac{1}{3}$. This distinction is vital for understanding exponential functions.

Visualizing the Decay: What the Graph Looks Like

Now that we've identified the key players – the initial value of 60 and the decay base of $\frac{1}{3}$ – let's paint a picture of what the graph of $f(x)=60\left(\frac{1}{3}\right)^x$ actually looks like. Imagine a standard coordinate plane with the x-axis running horizontally and the y-axis running vertically. We established that our graph starts at the point (0, 60). This is where the curve first meets the y-axis. Since the base is $\frac{1}{3}$, which is between 0 and 1, we know this is a decaying exponential function. This means as we move to the right along the x-axis (i.e., as $x$ increases), the corresponding y-values will decrease. The graph will descend from its starting point at (0, 60). Let's trace a few more points to solidify this:

• At $x=1$, $f(1) = 60 \times \left(\frac{1}{3}\right)^1 = 60 \times \frac{1}{3} = 20$. So, the point (1, 20) is on the graph.
• At $x=2$, $f(2) = 60 \times \left(\frac{1}{3}\right)^2 = 60 \times \frac{1}{9} = \frac{60}{9} = \frac{20}{3} \approx 6.67$. So, the point (2, 6.67) is on the graph.
• At $x=3$, $f(3) = 60 \times \left(\frac{1}{3}\right)^3 = 60 \times \frac{1}{27} = \frac{60}{27} = \frac{20}{9} \approx 2.22$. So, the point (3, 2.22) is on the graph.

As you can see, the y-values are getting smaller and smaller: 60, 20, 6.67, 2.22... The curve is dropping quite rapidly at first and then begins to level off as it gets closer to the x-axis. This behavior is known as asymptotic behavior. The x-axis (the line $y=0$) acts as a horizontal asymptote. This means the graph gets infinitely close to the x-axis but never actually touches or crosses it. The y-values approach 0 but never become 0 (or negative). Now, let's consider what happens as we move to the left of the y-axis (i.e., as $x$ decreases or becomes negative). Remember, $f(x) = 60 \times \left(\frac{1}{3}\right)^x$. When $x$ is negative, say $x=-1$:

• At $x=-1$, $f(-1) = 60 \times \left(\frac{1}{3}\right)^{-1} = 60 \times 3 = 180$. So, the point (-1, 180) is on the graph.
• At $x=-2$, f(-2) = 60 \times \left(\frac{1}{3} ight)^{-2} = 60 \times (3)^2 = 60 \times 9 = 540. So, the point (-2, 540) is on the graph.

As $x$ becomes more negative, the y-values increase dramatically. The graph shoots upwards steeply as we move to the left. So, to summarize the visual: the graph starts at (0, 60), decreases rapidly as $x$ increases, approaching the x-axis from above, and increases very rapidly as $x$ decreases, moving away from the x-axis. It's a smooth, continuous curve. Comparing this to the incorrect descriptions in the options: Option A and B both suggest an initial value of 20, which is incorrect. They also talk about subtracting $\frac{1}{3}$, which is the definition of an arithmetic sequence, not an exponential one. An exponential function is defined by multiplication (or division, which is multiplication by the reciprocal). Therefore, the correct description involves an initial value of 60 and a multiplicative factor (the base) of $\frac{1}{3}$ leading to exponential decay.

Addressing the Misconceptions in the Options

It's crucial to understand why the given options, A and B, are incorrect for the function $f(x)=60\left(\frac{1}{3}\right)^x$. These options reveal common misunderstandings about exponential functions, particularly the difference between exponential and arithmetic sequences, and the identification of the initial value. Let's break down option A: "The graph has an initial value of 20, and each successive term is determined by subtracting $\frac{1}{3}$." The first part, "initial value of 20," is incorrect. As we've thoroughly calculated, the initial value (when $x=0$) is $f(0) = 60\left(\frac{1}{3}\right)^0 = 60 \times 1 = 60$. The graph starts at y=60, not y=20. The second part, "each successive term is determined by subtracting $\frac{1}{3}$," describes an arithmetic sequence. An arithmetic sequence involves adding or subtracting a constant difference. If this were true, the function would look something like $g(x) = 20 - \frac{1}{3}x$. This is a linear function, not an exponential one. For an exponential function like f(x)=60\left(\frac{1}{3} ight)^x, the change between successive terms is multiplicative, not additive or subtractive. We multiply by the base, $\frac{1}{3}$. So, the statement should refer to multiplying by $\frac{1}{3}$, not subtracting it.

Now let's look at option B: "The graph has an initial value of 20..." This part is exactly the same error as in option A. The initial value is incorrectly stated as 20 when it should be 60. While option B doesn't explicitly state how successive terms are determined (it might have been cut off or intended to be similar to A), the initial value error alone makes it incorrect. If it were to say something like "multiplying by $\frac{1}{3}$" but still had the initial value wrong, it would still be fundamentally flawed. The structure $f(x) = a imes b^x$ clearly shows that '$a$' is the initial value (at $x=0$) and '$b$' is the base that determines the multiplicative change. The function f(x)=60\left(\frac{1}{3} ight)^x has $a=60$ and $b=\frac{1}{3}$. Therefore, any description that claims an initial value of 20 is definitively wrong. Furthermore, any description that suggests subtraction of the base is also wrong; it must be multiplication by the base. The correct description would state that the graph has an initial value of 60 and that each successive value is obtained by multiplying the previous value by $\frac{1}{3}$. This leads to exponential decay because the base $\frac{1}{3}$ is between 0 and 1. The graph begins at (0, 60) and approaches the x-axis as $x$ increases. It's important to pay close attention to both the coefficient and the base in exponential functions to accurately interpret their graphs. The options provided seem to confuse these fundamental aspects, leading to incorrect descriptions of the function's behavior.

Conclusion: The True Nature of $f(x)=60\left(\frac{1}{3}

ight)^x$

In conclusion, after a thorough examination of the function f(x)=60\left(\frac{1}{3} ight)^x, we can definitively state its graphical characteristics. The initial value of the function, which is the y-intercept, is found by evaluating $f(0)$. In this case, f(0) = 60\left(\frac{1}{3} ight)^0 = 60 \times 1 = 60. Therefore, the graph begins at the point (0, 60). This is a critical piece of information that immediately disqualifies any description suggesting an initial value of 20. The base of the exponential function is $\frac{1}{3}$. This base dictates the behavior of the function as $x$ changes. Since $\frac{1}{3}$ is between 0 and 1, the function exhibits exponential decay. This means that for every unit increase in $x$, the value of $f(x)$ is multiplied by $\frac{1}{3}$. The graph will decrease as $x$ increases, approaching the x-axis (y=0) as a horizontal asymptote. Conversely, as $x$ decreases (becomes more negative), the graph will increase rapidly. The common misconceptions presented in the options, such as an initial value of 20 or determining successive terms by subtraction, are incorrect because they misrepresent these fundamental properties of exponential functions. The correct description involves an initial value of 60 and a constant multiplicative factor of $\frac{1}{3}$ for each unit change in the exponent. Understanding these elements is paramount to correctly interpreting and visualizing the graph of any exponential function.

For further exploration into the fascinating world of exponential functions and their graphs, I recommend visiting Khan Academy's section on exponential functions. They offer excellent resources and practice problems that can deepen your understanding.