Exponential Functions: Finding Equations From Points
Have you ever looked at a graph that seems to shoot upwards or dwindle downwards at an incredible rate and wondered what kind of mathematical magic is at play? Chances are, you're looking at an exponential function! These functions are fundamental in describing phenomena that grow or decay rapidly, from population growth and compound interest to radioactive decay and the spread of viruses. Understanding how to write an exponential function for a graph, especially when you're given a couple of points it passes through, is a superpower in the world of mathematics. It allows you to model real-world situations, predict future trends, and solve complex problems. In this article, we'll dive deep into the process of crafting these powerful functions, breaking down how to find the equation of an exponential function given two points on its graph. We'll tackle several examples, transforming abstract coordinates into concrete mathematical models. So, buckle up, and let's get ready to explore the fascinating realm of exponential functions and unlock the secrets hidden within their graphs!
Understanding the General Form of an Exponential Function
Before we start plugging in numbers and solving for unknowns, it's crucial to get a firm grasp on the basic structure of an exponential function. The general form that we'll be working with is typically represented as y = abˣ. Here, 'a' and 'b' are constants that we need to determine, and 'x' and 'y' are our variables. The constant 'a' represents the initial value or the y-intercept (the value of y when x = 0). The constant 'b' is the base of the exponent, and it dictates the rate of growth or decay. If b > 1, the function exhibits exponential growth, meaning the y-values increase rapidly as x increases. If 0 < b < 1, the function exhibits exponential decay, where the y-values decrease rapidly as x increases. If b = 1, the function becomes a horizontal line y = a, which isn't truly exponential. If b ≤ 0, the function is generally not considered an exponential function in this standard form due to potential issues with domain and graph behavior. Our goal is to find the specific values of 'a' and 'b' that make this general form fit the points provided. This involves setting up a system of equations using the given coordinates and solving for 'a' and 'b'. It might sound a bit abstract now, but as we work through the examples, you'll see how this general form becomes our roadmap to finding the exact exponential function.
Solving for 'a' and 'b': A Step-by-Step Approach
Now that we know the basic blueprint, let's lay out the systematic steps to find the exponential function y = abˣ when given two points, say (x₁, y₁) and (x₂, y₂). The first step is to substitute each point into the general equation. This will give us two separate equations:
- For (x₁, y₁): y₁ = abˣ¹
- For (x₂, y₂): y₂ = abˣ²
We now have a system of two equations with two unknowns ('a' and 'b'). The most efficient way to solve this system is often by division. We can divide the second equation by the first equation (or vice versa). Let's divide equation (2) by equation (1):
(y₂ / y₁) = (abˣ²) / (abˣ¹)
Notice that the 'a' terms cancel out: (y₂ / y₁) = bˣ² / bˣ¹
Using the properties of exponents (when dividing powers with the same base, subtract the exponents), we get:
(y₂ / y₁) = b⁽ˣ²⁻ˣ¹⁾
Now, we need to solve for 'b'. To isolate 'b', we can raise both sides of the equation to the power of 1 / (x₂ - x₁):
b = (y₂ / y₁)^(1 / (x₂ - x₁))
Once you've calculated the value of 'b', you can substitute it back into either of the original two equations (y₁ = abˣ¹ or y₂ = abˣ²) to solve for 'a'. Let's use the first equation:
y₁ = abˣ¹
Divide both sides by bˣ¹ to isolate 'a':
a = y₁ / bˣ¹
After finding the values for both 'a' and 'b', you can plug them back into the general form y = abˣ to get your specific exponential function.
Example 1: Growth Through Points (2, 16) and (3, 32)
Let's put our steps into action with our first problem: finding the exponential function for a graph that passes through the points (2, 16) and (3, 32). We are looking for a function in the form y = abˣ.
Step 1: Substitute the points into the general equation.
For the point (2, 16): 16 = ab² (Equation 1)
For the point (3, 32): 32 = ab³ (Equation 2)
Step 2: Divide Equation 2 by Equation 1 to solve for 'b'.
(32 / 16) = (ab³) / (ab²)
2 = b³ / b²
2 = b⁽³⁻²⁾
2 = b¹
So, b = 2.
Step 3: Substitute the value of 'b' back into Equation 1 to solve for 'a'.
16 = a(2)²
16 = a(4)
Divide both sides by 4:
a = 16 / 4
a = 4
Step 4: Write the exponential function using the values of 'a' and 'b'.
With a = 4 and b = 2, our exponential function is: y = 4(2)ˣ
You can quickly check this by plugging the original points back in. For (2, 16): y = 4(2)² = 4(4) = 16. Correct! For (3, 32): y = 4(2)³ = 4(8) = 32. Correct! This function accurately models the graph passing through these points.
Example 2: Decay Through Points (1, 1) and (3, 0.25)
Now, let's tackle a problem that involves exponential decay. We need to find the exponential function for a graph that passes through (1, 1) and (3, 0.25). Again, we use the form y = abˣ.
Step 1: Substitute the points into the general equation.
For the point (1, 1): 1 = ab¹ or 1 = ab (Equation 1)
For the point (3, 0.25): 0.25 = ab³ (Equation 2)
Step 2: Divide Equation 2 by Equation 1 to solve for 'b'.
(0.25 / 1) = (ab³) / (ab)
0.25 = b³ / b¹
0.25 = b⁽³⁻¹⁾
0.25 = b²
To find 'b', we take the square root of both sides. Since we are dealing with exponential functions where the base 'b' is typically positive (0 < b < 1 for decay), we take the positive root:
b = √0.25
b = 0.5
Step 3: Substitute the value of 'b' back into Equation 1 to solve for 'a'.
1 = a(0.5)
Divide both sides by 0.5:
a = 1 / 0.5
a = 2
Step 4: Write the exponential function using the values of 'a' and 'b'.
With a = 2 and b = 0.5, our exponential function is: y = 2(0.5)ˣ
Let's check: For (1, 1): y = 2(0.5)¹ = 2(0.5) = 1. Correct! For (3, 0.25): y = 2(0.5)³ = 2(0.125) = 0.25. Correct! This function accurately models exponential decay through the given points.
Example 3: Larger Numbers Through Points (2, 90) and (4, 810)
This example involves slightly larger numbers, but the process remains the same. We want to find the exponential function for a graph passing through (2, 90) and (4, 810).
Step 1: Substitute the points into the general equation y = abˣ.
For the point (2, 90): 90 = ab² (Equation 1)
For the point (4, 810): 810 = ab⁴ (Equation 2)
Step 2: Divide Equation 2 by Equation 1 to solve for 'b'.
(810 / 90) = (ab⁴) / (ab²)
9 = b⁴ / b²
9 = b⁽⁴⁻²⁾
9 = b²
Taking the square root of both sides:
b = √9
b = 3
Step 3: Substitute the value of 'b' back into Equation 1 to solve for 'a'.
90 = a(3)²
90 = a(9)
Divide both sides by 9:
a = 90 / 9
a = 10
Step 4: Write the exponential function using the values of 'a' and 'b'.
With a = 10 and b = 3, our exponential function is: y = 10(3)ˣ
Check: For (2, 90): y = 10(3)² = 10(9) = 90. Correct! For (4, 810): y = 10(3)⁴ = 10(81) = 810. Correct! This function accurately models the graph through these points.
Example 4: Negative Exponents and Decay Through Points (-2, 4) and (1, 0.5)
This example introduces a negative x-coordinate, which can sometimes seem tricky, but the method is identical. We need to find the exponential function for a graph passing through (-2, 4) and (1, 0.5).
Step 1: Substitute the points into the general equation y = abˣ.
For the point (-2, 4): 4 = ab⁻² (Equation 1)
For the point (1, 0.5): 0.5 = ab¹ or 0.5 = ab (Equation 2)
Step 2: Divide Equation 2 by Equation 1 to solve for 'b'.
(0.5 / 4) = (ab) / (ab⁻²)
0.125 = b¹ / b⁻²
0.125 = b⁽¹⁻⁽⁻²⁾⁾
0.125 = b⁽¹⁺²⁾
0.125 = b³
To solve for 'b', we need to find the cube root of 0.125:
b = ³√0.125
b = 0.5
Step 3: Substitute the value of 'b' back into Equation 2 to solve for 'a'.
0.5 = a(0.5)¹
0.5 = a(0.5)
Divide both sides by 0.5:
a = 0.5 / 0.5
a = 1
Step 4: Write the exponential function using the values of 'a' and 'b'.
With a = 1 and b = 0.5, our exponential function is: y = 1(0.5)ˣ or simply y = (0.5)ˣ
Check: For (-2, 4): y = (0.5)⁻² = (1/2)⁻² = 2² = 4. Correct! For (1, 0.5): y = (0.5)¹ = 0.5. Correct! This function models the given points.
Example 5: Significant Growth Through Points (1, 12) and (3, 192)
Let's find the exponential function for a graph passing through (1, 12) and (3, 192).
Step 1: Substitute the points into the general equation y = abˣ.
For the point (1, 12): 12 = ab¹ or 12 = ab (Equation 1)
For the point (3, 192): 192 = ab³ (Equation 2)
Step 2: Divide Equation 2 by Equation 1 to solve for 'b'.
(192 / 12) = (ab³) / (ab)
16 = b³ / b¹
16 = b⁽³⁻¹⁾
16 = b²
Taking the square root of both sides:
b = √16
b = 4
Step 3: Substitute the value of 'b' back into Equation 1 to solve for 'a'.
12 = a(4)¹
12 = 4a
Divide both sides by 4:
a = 12 / 4
a = 3
Step 4: Write the exponential function using the values of 'a' and 'b'.
With a = 3 and b = 4, our exponential function is: y = 3(4)ˣ
Check: For (1, 12): y = 3(4)¹ = 3(4) = 12. Correct! For (3, 192): y = 3(4)³ = 3(64) = 192. Correct!
Example 6: Finding Decay Rate Through Points (1, 18) and (3, 72)
In this final example, we find the exponential function for a graph passing through (1, 18) and (3, 72).
Step 1: Substitute the points into the general equation y = abˣ.
For the point (1, 18): 18 = ab¹ or 18 = ab (Equation 1)
For the point (3, 72): 72 = ab³ (Equation 2)
Step 2: Divide Equation 2 by Equation 1 to solve for 'b'.
(72 / 18) = (ab³) / (ab)
4 = b³ / b¹
4 = b⁽³⁻¹⁾
4 = b²
Taking the square root of both sides:
b = √4
b = 2
Step 3: Substitute the value of 'b' back into Equation 1 to solve for 'a'.
18 = a(2)¹
18 = 2a
Divide both sides by 2:
a = 18 / 2
a = 9
Step 4: Write the exponential function using the values of 'a' and 'b'.
With a = 9 and b = 2, our exponential function is: y = 9(2)ˣ
Check: For (1, 18): y = 9(2)¹ = 9(2) = 18. Correct! For (3, 72): y = 9(2)³ = 9(8) = 72. Correct!
Conclusion: Mastering Exponential Functions
As we've journeyed through these examples, you've seen firsthand how to systematically derive an exponential function from just two points on its graph. The general form y = abˣ serves as our trusty blueprint, and by substituting our given points, we create a system of equations. The clever technique of dividing these equations allows us to isolate and solve for the base 'b', and then back-substituting helps us find the initial value 'a'. Whether the function represents rapid growth or steady decay, this method remains robust. Mastering this skill opens doors to understanding and modeling a vast array of real-world phenomena, from financial growth to biological processes. It's a fundamental concept in algebra that empowers you to translate graphical data into precise mathematical descriptions. Keep practicing, and you'll soon be writing exponential functions with confidence!
For further exploration into exponential functions and their applications, you can visit **Khan Academy **for in-depth lessons and practice exercises, or consult resources from **The Art of Problem Solving **for more advanced mathematical concepts.
