Exponential Function: Find Equation From Two Points

Oct 11, 2025 by ADMIN 52 views

Exponential Function: Finding the Equation from Two Points

Hey guys! Today, we're diving into the exciting world of exponential functions! Specifically, we're going to tackle a common problem: finding the equation of an exponential function when given two points it passes through. It might seem a bit tricky at first, but trust me, we'll break it down step by step, and you'll be a pro in no time. So, let's get started and learn how to find the exponential function that fits perfectly through any two given points!

Understanding Exponential Functions

Before we jump into solving the problem, let's make sure we're all on the same page about what an exponential function actually is. An exponential function has the general form:

y = a * b^x

Where:

y is the dependent variable (the output).
x is the independent variable (the input).
a is the initial value or the y-intercept (the value of y when x is 0).
b is the base, which is a positive number not equal to 1. This value determines whether the function represents exponential growth (if b > 1) or exponential decay (if 0 < b < 1).

The key characteristic of an exponential function is that the variable x appears as an exponent. This leads to rapid growth or decay as x changes. Understanding this form is crucial because our goal is to find the specific values of a and b that make the function pass through our given points.

Why Exponential Functions Matter

You might be wondering, "Why should I care about exponential functions?" Well, they pop up everywhere in the real world! From population growth and compound interest to radioactive decay and the spread of diseases, exponential functions are the perfect tool for modeling situations where quantities increase or decrease at a rate proportional to their current value.

For instance, think about a savings account with compound interest. The amount of money you earn each year isn't just a fixed amount; it's a percentage of your current balance. This leads to exponential growth, where your money grows faster and faster over time. Similarly, in biology, the number of bacteria in a culture can double every hour, leading to an exponential increase in population. So, mastering exponential functions isn't just about math class; it's about understanding the world around you!

Setting Up the Equations

Okay, now that we've refreshed our understanding of exponential functions, let's tackle the problem at hand. We're given two points: (-2, 7/25) and (2, 175). Our mission is to find the specific exponential function in the form y = a * b^x that passes through both of these points. How do we do it? By using the information these points give us to create a system of equations!

Each point (x, y) gives us a pair of values that satisfy the exponential function's equation. So, we can plug the x and y values from each point into our general equation, creating two separate equations. Let's do that:

Point 1: (-2, 7/25) Substituting x = -2 and y = 7/25 into y = a * b^x, we get:
```
7/25 = a * b^(-2)
```
Point 2: (2, 175) Substituting x = 2 and y = 175 into y = a * b^x, we get:
```
175 = a * b^(2)
```

Now we have two equations with two unknowns (a and b). This is a classic system of equations problem! Solving this system will give us the values of a and b that define our exponential function. We're one step closer to cracking the code!

Why Two Points?

You might be wondering, "Why do we need two points?" Well, think of it like this: an exponential function has two key parameters that define its shape – the initial value (a) and the base (b). Each point on the graph of the function provides a piece of information, a constraint that the function must satisfy. One point alone isn't enough to nail down both a and b. It's like trying to draw a line with only one point – you can draw infinitely many lines through that single point!

However, with two points, we have two independent pieces of information. This allows us to create a system of two equations, which we can then solve to find the unique values of a and b that define the exponential function passing through those points. It's like having two dots on a piece of paper – there's only one line that can connect them!

Solving the System of Equations

Alright, we've got our system of equations:

7/25 = a * b^(-2)
175 = a * b^(2)

Now comes the fun part: solving for a and b. There are a couple of ways we can approach this, but a common and efficient method is to use substitution. The idea behind substitution is to solve one equation for one variable and then substitute that expression into the other equation. This eliminates one variable, leaving us with a single equation that we can solve for the remaining variable.

Let's solve equation (1) for a. To do this, we'll multiply both sides of the equation by b^2:

(7/25) * b^2 = a * b^(-2) * b^2
(7/25) * b^2 = a

So, we have a = (7/25) * b^2. Now, we'll substitute this expression for a into equation (2):

175 = ((7/25) * b^2) * b^2

This simplifies to:

175 = (7/25) * b^4

Now we have a single equation with just one variable, b. To solve for b, let's first multiply both sides by 25/7:

175 * (25/7) = b^4
625 = b^4

To find b, we need to take the fourth root of both sides:

b = ±√(√625) = ±5

Since the base of an exponential function must be positive, we take b = 5. Great! We've found b!

Why Substitution Works

Substitution is a powerful technique for solving systems of equations because it allows us to reduce the problem from multiple variables to a single variable. It's like having a complex puzzle with many pieces – substitution helps us isolate one piece, solve it, and then use that solution to figure out the rest of the puzzle.

In our case, by solving one equation for a in terms of b, we were able to eliminate a from the second equation. This left us with an equation involving only b, which we could then solve directly. Once we found b, we could easily plug it back into our expression for a to find its value. It's a clever way to break down a seemingly difficult problem into smaller, more manageable steps.

Finding the Value of 'a'

Now that we've found b = 5, we're halfway there! Remember, our goal is to determine the exponential function y = a * b^x, so we still need to find the value of a. Luckily, we've already done the hard work. We have an expression for a in terms of b: a = (7/25) * b^2.

So, all we need to do is plug in our value of b into this equation:

a = (7/25) * (5)^2
a = (7/25) * 25
a = 7

Fantastic! We've found a = 7. Now we have both a and b, which means we can write the complete exponential function.

The Importance of Back-Substitution

Finding b is a major step, but it's crucial to remember that we're not done until we find both a and b. This is where back-substitution comes in. Back-substitution simply means plugging the value we found for one variable (in this case, b) back into one of our original equations (or an equation derived from them) to solve for the other variable (a).

It's like building a bridge – finding b is like laying the first cornerstone, but we need to lay the other cornerstones (find a) to complete the structure and make it strong. Back-substitution ensures that we use all the information we have to find the most accurate and complete solution.

Writing the Exponential Function

We've done it! We've found the values of a and b that define the exponential function passing through the points (-2, 7/25) and (2, 175). We have:

a = 7
b = 5

Now, we simply plug these values into the general form of an exponential function, y = a * b^x:

y = 7 * 5^x

And there you have it! This is the exponential function that passes through the given points. We've successfully navigated the problem from start to finish!

Double-Checking Our Work

It's always a good idea to double-check our work, especially in math. We can do this by plugging our original points (-2, 7/25) and (2, 175) back into our exponential function y = 7 * 5^x to see if they satisfy the equation.

For (-2, 7/25):
```
y = 7 * 5^(-2)
y = 7 * (1/25)
y = 7/25
```
This checks out!
For (2, 175):
```
y = 7 * 5^(2)
y = 7 * 25
y = 175
```
This also checks out!

Since our function satisfies both points, we can be confident that our answer is correct. This step is a great way to catch any potential errors and ensure that our solution is accurate.

Conclusion

Woohoo! We've successfully found the exponential function that passes through the points (-2, 7/25) and (2, 175). We did this by setting up a system of equations, using substitution to solve for the base b, and then plugging that value back in to find the initial value a. Finally, we wrote the equation and double-checked our work to make sure everything was perfect.

Remember, the key to mastering exponential functions is understanding their general form, knowing how to set up equations from given points, and being comfortable with algebraic manipulation. With practice, you'll become a pro at solving these types of problems. Keep up the great work, guys, and keep exploring the fascinating world of mathematics!