Exponential Functions: Finding The Right Equation

Hey everyone! Today, we're diving into the world of exponential functions. We'll tackle a common type of math problem: finding the equation of an exponential function that goes through a specific point. Don't worry, it's not as scary as it sounds! We'll break it down step-by-step, making sure you grasp the concepts, so you can ace similar problems. So, if you've ever wondered how to nail down the right exponential function, stick around – this is for you! This problem focuses on identifying an exponential function that passes through a particular point, specifically (2, 36). To solve this type of problem, we need to understand the characteristics of exponential functions and how to substitute values to find the correct equation. In this case, we have a point (2, 36), which means when x = 2, f(x) (or y) = 36. We are given a set of equations, and our job is to figure out which one fits this condition. The key here is substituting the x-value (2 in this case) into each equation and seeing which one produces an output (f(x) or y) of 36. Let's start with a quick refresher on what an exponential function actually is. An exponential function is a function of the form f(x) = a * b^x, where 'a' is the initial value, 'b' is the base (a positive number not equal to 1), and x is the exponent. The base b determines the rate of growth or decay. If b > 1, the function grows exponentially; if 0 < b < 1, it decays exponentially. Now, let’s explore our options and do some calculations. We'll go through each of the given choices, substituting x = 2, and see which one yields an f(x) value of 36.

Decoding the Problem: What's an Exponential Function?

Alright, before we get to the juicy part – solving the problem – let's make sure we're all on the same page about what an exponential function actually is. Think of it like this: exponential functions are all about rapid growth or decay. Unlike linear functions (where things change at a constant rate), exponential functions change by a factor. The general form of an exponential function is f(x) = a * b^x. Where, a is the initial value, b is the base and x is the exponent. The base, b, is super important! It determines whether the function increases (grows) or decreases (decays) as x increases. If b is greater than 1, you've got exponential growth – think of compound interest or the spread of a virus. If b is between 0 and 1, you've got exponential decay – think of radioactive decay or the depreciation of a car. Now, in the problem, we're given a specific point (2, 36), meaning when x = 2, the function's output, f(x) or y, equals 36. Our mission is to find the function that satisfies this condition. This means we'll plug in x = 2 into each of the equations and see which one gives us f(x) = 36. This process is called evaluating the function at a specific point. Remember that the base, b, must be positive and cannot be equal to 1. This restriction is crucial because if b = 1, the function would simply be a constant function (y = a), not an exponential one. Understanding this setup is fundamental to solving the problem. So, are you ready to dive into the equations and find our answer? Let’s do it!

Solving Step-by-Step: Testing the Equations

Alright, guys, let's roll up our sleeves and get to the core of the problem: testing the equations. We have a set of potential exponential function equations. Our goal is to figure out which one passes through the point (2, 36). The key is simple: substitute x = 2 into each equation and see if you get f(x) = 36. Let’s go through each option methodically:

Option A: f(x) = 4 * 3^x

Let’s start with option A: f(x) = 4 * 3^x. To test this, we substitute x = 2: f(2) = 4 * 3^2. First, we calculate the exponent: 3^2 = 9. Then, we multiply by 4: f(2) = 4 * 9 = 36. Hey, we've found our match! When x = 2, f(x) = 36. This suggests that the point (2, 36) does indeed lie on the graph of this function. However, we'll continue to examine the other options to be absolutely certain.

Option B: f(x) = 4 * x^3

Next up, option B: f(x) = 4 * x^3. Here, we substitute x = 2: f(2) = 4 * 2^3. First, calculate the exponent: 2^3 = 8. Then, multiply by 4: f(2) = 4 * 8 = 32. Since f(2) is 32, not 36, this function does not pass through the point (2, 36). We can eliminate this option.

Option C: f(x) = 6 * 3^x

Let's test option C: f(x) = 6 * 3^x. Substitute x = 2: f(2) = 6 * 3^2. Calculate the exponent first: 3^2 = 9. Then, multiply by 6: f(2) = 6 * 9 = 54. The result is 54, which doesn’t match our target of 36. Therefore, this function does not pass through the point (2, 36).

Option D: f(x) = 6 * x^3

Finally, let's check option D: f(x) = 6 * x^3. Substitute x = 2: f(2) = 6 * 2^3. Calculate the exponent: 2^3 = 8. Multiply by 6: f(2) = 6 * 8 = 48. This is also not 36, so this equation does not represent the function we are looking for.

The Verdict: Which Equation Works?

So, after all that work, which equation fits the bill? Let's recap. We went through each of the four provided equations, substituting x = 2 and calculating f(2) for each. Here's what we found:

• Option A: f(x) = 4 * 3^x. When x = 2, f(2) = 36. Bingo! This is the correct equation.
• Option B: f(x) = 4 * x^3. When x = 2, f(2) = 32. Not the right one.
• Option C: f(x) = 6 * 3^x. When x = 2, f(2) = 54. Nope!
• Option D: f(x) = 6 * x^3. When x = 2, f(2) = 48. Not correct.

As you can see, only option A, f(x) = 4 * 3^x, gives us the desired output of 36 when x = 2. This means the graph of this exponential function actually passes through the point (2, 36). The base of the exponential function is 3, indicating exponential growth. The constant 4 scales the function, determining the initial value adjusted for the exponential change. This kind of problem is typical in math, especially in algebra and precalculus. It helps you understand how different parameters of an exponential function affect its output and its behavior on a graph. By systematically testing each option, you can quickly find the one that fits your given point. This approach is not only helpful for this specific problem, but also provides a useful method for solving a broad range of related problems involving functions and their graphs. So, keep practicing, keep learning, and you'll become a pro at these problems in no time! Keep up the great work, everyone!