Reflecting Exponential Functions Across The X-Axis A Detailed Explanation

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In the world of mathematics, understanding transformations of functions is crucial for grasping their behavior and applications. Among these transformations, reflections hold a significant place. When a function is reflected across the x-axis, its graphical representation is flipped vertically. This concept is particularly interesting when applied to exponential functions, which exhibit rapid growth or decay. In this comprehensive guide, we will delve into the reflection of the exponential function f(x)=2(3.5)xf(x) = 2(3.5)^x across the x-axis, exploring how this transformation affects its equation and initial value. By the end of this guide, you will have a solid understanding of how reflections work and how they impact the characteristics of exponential functions.

Let's begin by understanding the basics of exponential functions. An exponential function is defined as f(x)=abxf(x) = ab^x, where aa is the initial value and bb is the base. The base bb determines the growth or decay rate of the function. If b>1b > 1, the function exhibits exponential growth, and if 0<b<10 < b < 1, it exhibits exponential decay. In our case, the given function is f(x)=2(3.5)xf(x) = 2(3.5)^x. Here, the initial value aa is 2, and the base bb is 3.5. Since 3. 5 > 1, this function represents exponential growth. Now, let's focus on what happens when we reflect this function across the x-axis. Reflection across the x-axis means that for every point (x,y)(x, y) on the original function, the corresponding point on the reflected function will be (x,βˆ’y)(x, -y). In simpler terms, the y-coordinate changes its sign while the x-coordinate remains the same. Mathematically, if we have a function f(x)f(x), its reflection across the x-axis is given by g(x)=βˆ’f(x)g(x) = -f(x). This is a crucial concept to remember as we move forward. Applying this to our function f(x)=2(3.5)xf(x) = 2(3.5)^x, we find the reflected function g(x)g(x) by multiplying the entire function by -1. This gives us g(x)=βˆ’2(3.5)xg(x) = -2(3.5)^x. Notice how the only change is the sign of the initial value. This negative sign is what causes the reflection across the x-axis. The base, 3.5, remains the same, indicating that the rate of growth or decay is still determined by this value. However, the negative sign in front of the 2 signifies that the function is now reflected, meaning it will decrease as x increases, unlike the original function which increased. Understanding this simple yet powerful transformation helps in visualizing and analyzing the behavior of exponential functions under different conditions. It’s not just about changing a sign; it’s about altering the entire direction of the function's graph.

Determining the Reflected Function g(x)

To determine the function definition of g(x)g(x), which is the reflection of f(x)=2(3.5)xf(x) = 2(3.5)^x across the x-axis, we need to apply the principle that reflecting a function across the x-axis involves negating the entire function. In other words, if f(x)f(x) represents the original function, then its reflection g(x)g(x) is given by g(x)=βˆ’f(x)g(x) = -f(x). This is a fundamental concept in transformations of functions, and it's essential to grasp this principle to correctly manipulate functions and their graphs. In our case, the original function is f(x)=2(3.5)xf(x) = 2(3.5)^x. To find g(x)g(x), we multiply the entire expression by -1. This means we take the entire term 2(3.5)x2(3.5)^x and change its sign. Therefore, g(x)=βˆ’1βˆ—[2(3.5)x]g(x) = -1 * [2(3.5)^x]. Multiplying this out, we get g(x)=βˆ’2(3.5)xg(x) = -2(3.5)^x. This is the function definition of g(x)g(x), which represents the reflection of f(x)f(x) across the x-axis. Notice that the only change is the sign of the coefficient 2. The base of the exponential term, 3.5, and the variable xx remain the same. This is because the reflection across the x-axis only affects the vertical position of the graph, not its horizontal scaling or shape. The negative sign in front of the 2 indicates that the graph of g(x)g(x) will be a mirror image of f(x)f(x) across the x-axis. Where f(x)f(x) was positive, g(x)g(x) will be negative, and vice versa. This is a direct consequence of the reflection. Now, let's delve deeper into the implications of this reflection. The original function f(x)=2(3.5)xf(x) = 2(3.5)^x is an increasing exponential function because its base, 3.5, is greater than 1. As xx increases, the value of f(x)f(x) also increases, and the graph rises exponentially. However, when we reflect this function across the x-axis to obtain g(x)=βˆ’2(3.5)xg(x) = -2(3.5)^x, the function becomes a decreasing exponential function. This is because the negative sign in front of the 2 effectively flips the graph vertically. As xx increases, the value of g(x)g(x) becomes more and more negative, and the graph descends exponentially. This change in behavior is a critical aspect of understanding reflections. It’s not just a visual flip; it changes the fundamental nature of the function's growth or decay. This understanding is crucial when applying exponential functions to real-world scenarios, such as modeling population growth, radioactive decay, or financial investments. A reflection can completely alter the outcome, so it's essential to recognize and account for these transformations.

Identifying the Initial Value of g(x)

After determining the reflected function g(x)=βˆ’2(3.5)xg(x) = -2(3.5)^x, the next important step is to identify its initial value. The initial value of a function is the value of the function when x=0x = 0. It represents the starting point of the function's graph on the y-axis. In the general form of an exponential function, f(x)=abxf(x) = ab^x, the initial value is represented by the coefficient aa. This is because when x=0x = 0, b0=1b^0 = 1, and therefore f(0)=aβˆ—1=af(0) = a * 1 = a. In our case, the reflected function is g(x)=βˆ’2(3.5)xg(x) = -2(3.5)^x. To find the initial value of g(x)g(x), we need to evaluate g(0)g(0). This means we substitute x=0x = 0 into the equation: g(0)=βˆ’2(3.5)0g(0) = -2(3.5)^0. Now, we know that any non-zero number raised to the power of 0 is 1. Therefore, (3.5)0=1(3.5)^0 = 1. Substituting this back into the equation, we get: g(0)=βˆ’2βˆ—1=βˆ’2g(0) = -2 * 1 = -2. Thus, the initial value of g(x)g(x) is -2. This means that the graph of g(x)g(x) starts at the point (0, -2) on the y-axis. This is a crucial piece of information for understanding the behavior of the function. It tells us where the function begins its exponential descent. The initial value is particularly significant in real-world applications of exponential functions. For example, if g(x)g(x) represents the amount of a radioactive substance remaining after xx years, then the initial value -2 (in this mathematical context, we'd consider the absolute value 2 as the initial amount) would represent the initial amount of the substance. Similarly, if g(x)g(x) represents the value of an investment that is decreasing exponentially, the initial value would represent the starting value of the investment. In the context of reflections, the initial value of the reflected function is the negative of the initial value of the original function. This is because the reflection across the x-axis flips the graph vertically, changing the sign of the y-coordinate at every point, including the initial point where x=0x = 0. Therefore, if the original function f(x)f(x) had an initial value of 2, its reflection g(x)g(x) will have an initial value of -2. This relationship between the initial values of a function and its reflection is a direct consequence of the transformation and is a key concept to remember when working with reflections.

Conclusion

In summary, reflecting the function f(x)=2(3.5)xf(x) = 2(3.5)^x across the x-axis results in the function g(x)=βˆ’2(3.5)xg(x) = -2(3.5)^x. The initial value of g(x)g(x) is -2. This transformation involves negating the entire function, which changes the sign of the initial value and causes the graph to flip vertically across the x-axis. Understanding reflections and their impact on functions is essential for mathematical analysis and real-world applications. By grasping these concepts, you can effectively manipulate and interpret functions in various contexts.

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