Analyzing Exponential Function H(x) = (1/4)^x - 4096

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In this article, we delve into the fascinating world of exponential functions by closely examining the function H(x) = (1/4)^x - 4096. Our primary goal is to explore the behavior of this function by addressing two key questions. First, we aim to determine the value of H(-5) and identify the corresponding point on the graph of H. Second, we seek to find the value of x for which H(x) equals -3840, and subsequently, pinpoint the associated point on the graph. This exploration will not only enhance our understanding of exponential functions but also showcase how to evaluate them at specific points and solve for input values given a particular output. By meticulously working through these problems, we will gain valuable insights into the properties and characteristics of exponential functions, strengthening our mathematical foundation and problem-solving skills. This analysis is crucial for anyone seeking a deeper understanding of mathematical functions and their applications in various fields.

The first part of our exploration involves calculating the value of H(-5) for the exponential function H(x) = (1/4)^x - 4096. To accomplish this, we will substitute -5 for x in the function's equation. This substitution allows us to determine the output value of the function when the input is -5. The process of evaluation is fundamental in understanding how a function behaves at different points in its domain. By performing this calculation, we gain a concrete understanding of the function's value at a specific input, which helps us to visualize its behavior and characteristics. Moreover, this process reinforces our understanding of exponential functions and how they respond to negative exponents. The result of this evaluation will not only provide a numerical value but also enable us to identify a specific point on the graph of the function. This point, represented as a coordinate pair (x, H(x)), will be crucial in visualizing the function's curve and its position within the coordinate plane.

Step-by-step evaluation of H(-5):

  1. Substitute -5 for x in the function: H(-5) = (1/4)^(-5) - 4096
  2. Recall that a negative exponent means taking the reciprocal of the base raised to the positive exponent: (1/4)^(-5) = 4^5
  3. Calculate 4^5: 4^5 = 4 * 4 * 4 * 4 * 4 = 1024
  4. Substitute the result back into the equation: H(-5) = 1024 - 4096
  5. Perform the subtraction: H(-5) = -3072

Therefore, H(-5) = -3072. This result tells us that when the input to the function H(x) is -5, the output is -3072. This is a critical piece of information that allows us to understand the function's behavior at this particular point. The negative output value indicates that the function's graph lies below the x-axis at x = -5. This understanding is crucial for visualizing the overall shape and position of the graph of H(x). Moreover, this calculation demonstrates the significant impact of exponential functions, where a small change in the input can lead to a substantial change in the output. This characteristic is fundamental to the nature of exponential growth and decay, which are concepts widely applied in various fields such as finance, biology, and physics. The precise calculation of H(-5) is a cornerstone for further analysis and interpretation of the function's properties.

Now that we have determined H(-5) = -3072, we can identify the corresponding point on the graph of H. A point on a graph is represented by an ordered pair (x, y), where x is the input value and y is the output value. In this case, our input value is -5, and our output value is -3072. Therefore, the point that corresponds to H(-5) on the graph of H is (-5, -3072). This point represents a specific location on the graph, indicating where the function's curve passes when x equals -5. Understanding how to identify and plot points on a graph is essential for visualizing and interpreting the behavior of functions. The point (-5, -3072) provides a concrete visual anchor for the function H(x), allowing us to imagine its position in the coordinate plane. This is particularly useful for exponential functions, which can have rapidly changing values. By plotting this point, we gain a better sense of the function's overall trend and curvature, which is crucial for predicting its behavior at other points and for understanding its applications in various contexts. The ability to connect algebraic calculations with graphical representations is a fundamental skill in mathematics and provides a powerful tool for problem-solving and analysis.

In the second part of our exploration, we aim to determine the value of x for which H(x) = -3840. This involves solving the equation (1/4)^x - 4096 = -3840. Unlike the previous part, where we evaluated the function at a specific x value, here we are working in reverse. We are given the output of the function and asked to find the corresponding input. This type of problem-solving is a fundamental skill in algebra and calculus, as it allows us to understand the inverse relationship between inputs and outputs in a function. Solving for x in this context requires us to manipulate the equation algebraically, isolating x on one side. This process often involves undoing the operations performed by the function, such as exponentiation and subtraction. The techniques used to solve this equation, including adding constants to both sides and using logarithms, are crucial tools in mathematical analysis and are applicable in a wide range of problems. The solution we find will provide us with a specific x value that, when plugged into the function H(x), results in an output of -3840. This value is essential for fully understanding the function's behavior and its graphical representation.

Step-by-step solution for x when H(x) = -3840:

  1. Set H(x) equal to -3840: (1/4)^x - 4096 = -3840
  2. Add 4096 to both sides of the equation: (1/4)^x = -3840 + 4096
  3. Simplify the right side: (1/4)^x = 256
  4. Express 256 as a power of 4: 256 = 4^4
  5. Rewrite the equation using the fact that 1/4 = 4^(-1): (4(-1))x = 4^4
  6. Simplify the left side using the power of a power rule: 4^(-x) = 4^4
  7. Since the bases are equal, the exponents must be equal: -x = 4
  8. Multiply both sides by -1 to solve for x: x = -4

Therefore, x = -4 is the solution. This means that when the input to the function H(x) is -4, the output is -3840. This result is crucial for understanding the specific behavior of the function and its graph. The value of x = -4 provides a precise point at which the function H(x) achieves a particular output, allowing us to further analyze its properties. This calculation demonstrates the importance of algebraic manipulation in solving equations and understanding the relationships between variables. The ability to solve for x in equations like this is fundamental to many mathematical and scientific applications, including modeling exponential growth and decay, analyzing financial data, and understanding physical phenomena.

Now that we have found x = -4 when H(x) = -3840, we can identify the corresponding point on the graph of H. Similar to part (a), a point on a graph is represented by an ordered pair (x, y), where x is the input value and y is the output value. In this case, our input value is -4, and our output value is -3840. Therefore, the point that corresponds to H(x) = -3840 on the graph of H is (-4, -3840). This point represents another specific location on the graph of the function, indicating where the curve passes when H(x) equals -3840. By identifying this point, we gain further insight into the function's behavior and its graphical representation. The point (-4, -3840) provides a visual reference for the function's position in the coordinate plane, reinforcing our understanding of its overall trend and curvature. This ability to connect algebraic solutions with graphical representations is a powerful tool in mathematics, enabling us to visualize abstract concepts and solve complex problems. By plotting this point alongside the one we found in part (a), we can begin to sketch the graph of H(x) and predict its behavior at other points, further enhancing our understanding of exponential functions and their applications.

In conclusion, we have successfully explored the exponential function H(x) = (1/4)^x - 4096 by addressing two key questions. First, we calculated H(-5) = -3072 and identified the corresponding point on the graph as (-5, -3072). This allowed us to understand the function's output for a specific negative input value and visualize its position in the coordinate plane. Second, we solved for x when H(x) = -3840, finding that x = -4, and identified the corresponding point as (-4, -3840). This demonstrated our ability to work in reverse, finding the input value given a specific output, and further enhanced our understanding of the function's behavior. By solving these two problems, we have not only strengthened our algebraic and problem-solving skills but also gained valuable insights into the properties and characteristics of exponential functions. These skills are crucial for understanding more advanced mathematical concepts and for applying these functions in various real-world contexts. The points we identified on the graph of H(x) provide concrete visual anchors for understanding its shape and position, highlighting the connection between algebraic calculations and graphical representations. This exploration serves as a solid foundation for further study of exponential functions and their applications in mathematics, science, and engineering.