Finding Exponential Function Formula Passing Through Points (-1, 5/4) And (2, 80)

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Finding the formula for an exponential function that passes through given points involves a systematic approach using the properties of exponential functions and algebraic manipulation. In this comprehensive guide, we will explore the step-by-step process of determining the equation of an exponential function that satisfies specific conditions, providing clarity and insights into this fundamental concept in mathematics. Understanding exponential functions is crucial in various fields, including finance, biology, and physics, where exponential growth and decay models are widely used. The ability to derive the formula for an exponential function empowers us to analyze and predict the behavior of systems that exhibit exponential patterns. This exploration will enhance your understanding of mathematical modeling and its practical applications. Let's dive into the methods and techniques required to find the formula for an exponential function through specified points, ensuring you grasp every detail for a solid foundation in this topic.

Understanding Exponential Functions

Before diving into the calculation, it's essential to understand the basic form of an exponential function. An exponential function is generally represented as y = ab^x, where:

  • y is the dependent variable.
  • a is the initial value or the y-intercept (the value of y when x is 0).
  • b is the base, which determines the rate of growth or decay. If b > 1, the function represents exponential growth; if 0 < b < 1, it represents exponential decay.
  • x is the independent variable.

Our goal is to find the values of a and b that make the function pass through the given points. This process involves using the coordinates of the points to create a system of equations, which we can then solve to find the values of a and b. The base b is particularly important as it dictates the nature of the exponential change – whether it is rapidly increasing or decreasing. The initial value a scales the function vertically and is crucial for accurately modeling real-world scenarios. By understanding these components, we can effectively construct and interpret exponential functions for a variety of applications.

Step 1: Setting Up the Equations

Given two points $\left(-1, \frac{5}{4}\right)$ and $(2, 80)$, we can substitute these into the general form of the exponential function y = ab^x to create two equations.

For the point $\left(-1, \frac{5}{4}\right)$, we have:

54=ab−1\frac{5}{4} = ab^{-1}

For the point $(2, 80)$, we have:

80=ab280 = ab^2

These two equations form a system of equations that we can solve for a and b. The first equation represents the function's behavior at x = -1, and the second equation describes it at x = 2. By using these points, we effectively anchor the exponential curve to these specific locations, allowing us to determine the parameters that define the curve's shape and position. This method is a fundamental application of algebra in the context of exponential functions, providing a concrete way to translate graphical information into algebraic equations.

Step 2: Solving for a in Terms of b

To solve this system, we can first isolate a in one of the equations. Let's use the first equation:

54=ab−1\frac{5}{4} = ab^{-1}

Multiply both sides by b to solve for a:

a=54ba = \frac{5}{4}b

Now we have an expression for a in terms of b. This step is crucial because it allows us to reduce the system of two equations with two variables into a single equation with one variable. By expressing a as a function of b, we can substitute this expression into the second equation, thereby eliminating a and making it possible to solve for b. This algebraic manipulation is a common technique in solving systems of equations and highlights the interconnectedness of the variables in the exponential function.

Step 3: Substituting and Solving for b

Now, substitute the expression for a into the second equation:

80=ab280 = ab^2

80=(54b)b280 = \left(\frac{5}{4}b\right)b^2

Simplify the equation:

80=54b380 = \frac{5}{4}b^3

Multiply both sides by $\frac{4}{5}$:

b3=80â‹…45b^3 = 80 \cdot \frac{4}{5}

b3=64b^3 = 64

Take the cube root of both sides:

b=643b = \sqrt[3]{64}

b=4b = 4

So, we have found the value of b, which is 4. This base value indicates that the function represents exponential growth, as b is greater than 1. The process of solving for b involved several algebraic steps, including substitution, simplification, and taking the cube root. Each step is crucial to isolating b and finding its value accurately. This part of the solution demonstrates the power of algebraic techniques in unraveling the parameters of an exponential function.

Step 4: Solving for a

Now that we have the value of b, we can substitute it back into the expression for a:

a=54ba = \frac{5}{4}b

a=54(4)a = \frac{5}{4}(4)

a=5a = 5

So, the value of a is 5. This a value represents the initial value of the exponential function, which is the y-intercept. Substituting the value of b into the equation we derived earlier allows us to directly calculate a without needing to solve another system of equations. This step highlights the efficiency of using intermediate results to find other unknowns in a mathematical problem. With both a and b determined, we now have all the components needed to write the complete exponential function.

Step 5: Writing the Exponential Function

Now that we have found a = 5 and b = 4, we can write the exponential function:

y=abxy = ab^x

y=5(4)xy = 5(4)^x

This is the exponential function that passes through the points $\left(-1, \frac{5}{4}\right)$ and $(2, 80)$. This final step is the culmination of all the previous steps, where we synthesize the values of a and b into the general form of the exponential function. The resulting equation, y = 5(4)^x, completely describes the exponential relationship that satisfies the given conditions. It provides a concise mathematical model that can be used to predict the value of y for any given x, demonstrating the practical utility of finding exponential functions through specific points.

Conclusion

Finding the formula for an exponential function that passes through specific points involves setting up a system of equations, solving for the base b, and then finding the initial value a. By following these steps, we can accurately determine the exponential function that fits given data points. The process of finding the exponential function through given points is a foundational skill in mathematics with wide-ranging applications. By systematically setting up and solving equations, we can derive precise mathematical models that describe exponential relationships. This method not only enhances our understanding of exponential functions but also provides a valuable tool for analyzing and predicting real-world phenomena. Whether in finance, biology, or other fields, the ability to determine exponential functions is essential for effective modeling and decision-making.

Final Answer: The final answer is $\boxed{y=5(4)^x}$