How To Find Equation Of Exponential Graph With Two Points

How to Find the Equation of an Exponential Graph with Two Points

Finding the equation of an exponential graph given two points is a common problem in algebra and calculus. The general form of an exponential equation is y = ab^x, where 'a' is the initial value (y-intercept when x=0) and 'b' is the base, representing the constant multiplicative factor. This article will guide you through the process, offering multiple approaches and examples to solidify your understanding.

Understanding the Exponential Function

Before diving into the methods, let's review the key characteristics of exponential functions:

• Growth or Decay: The value of 'b' determines whether the graph represents exponential growth (b > 1) or decay (0 < b < 1).
• Y-intercept: The y-intercept is the point where the graph crosses the y-axis (x=0). In the equation y = ab^x, the y-intercept is 'a'.
• Asymptote: Exponential functions have a horizontal asymptote, typically the x-axis (y=0), meaning the graph approaches but never touches this line.

Method 1: Using Simultaneous Equations

This is the most direct method, involving setting up and solving a system of two equations. Let's assume we have two points (x₁, y₁) and (x₂, y₂). We can substitute these points into the general equation y = ab^x to create two equations:

• y₁ = ab^x₁
• y₂ = ab^x₂

Steps:

1. Divide the Equations: Divide the second equation by the first equation. This eliminates 'a', leaving an equation solely in terms of 'b':

   y₂/y₁ = (ab^x₂)/(ab^x₁) = b^{x₂ - x₁}

2. Solve for 'b': Solve the resulting equation for 'b' by taking the logarithm (usually base 10 or natural logarithm, ln) of both sides:

   log(y₂/y₁) = (x₂ - x₁)log(b) log(b) = log(y₂/y₁) / (x₂ - x₁) b = 10^{[log(y₂/y₁) / (x₂ - x₁)]} (if using base 10 logarithm) b = e^{[ln(y₂/y₁) / (x₂ - x₁)]} (if using natural logarithm)

3. Solve for 'a': Substitute the value of 'b' back into either of the original equations (y₁ = ab^x₁ or y₂ = ab^x₂) and solve for 'a'.

4. Write the Equation: Finally, substitute the values of 'a' and 'b' into the general equation y = ab^x to obtain the equation of the exponential graph.

Example:

Let's find the equation of the exponential graph passing through the points (1, 6) and (3, 24).

1. Substitute the points:

   6 = ab¹ 24 = ab³

2. Divide the equations:

   24/6 = (ab³)/(ab¹) 4 = b²

3. Solve for 'b':

   b = √4 = 2 (We choose the positive root since the base of an exponential function is typically positive)

4. Solve for 'a': Substitute b = 2 into 6 = ab¹:

   6 = a(2)¹ a = 3

5. Write the equation:

   The equation of the exponential graph is y = 3(2)^x

Method 2: Using Logarithmic Transformation

This method involves transforming the exponential equation into a linear equation using logarithms. This allows us to use linear regression techniques or simpler methods to find 'a' and 'b'.

Steps:

1. Take the Logarithm: Take the natural logarithm (ln) of both sides of the equation y = ab^x:

   ln(y) = ln(ab^x) = ln(a) + xln(b)

2. Linear Transformation: Let Y = ln(y), A = ln(a), and B = ln(b). The equation becomes:

   Y = A + Bx

This is now a linear equation in the form Y = mX + c, where m = B and c = A.

3. Linear Regression (or use two points): If you have more than two points, use linear regression techniques to find the best-fit line (which gives the values of A and B). If you have only two points, you can use the two-point form of a line:

   (Y₂ - Y₁) / (x₂ - x₁) = B A = Y₁ - Bx₁ (or A = Y₂ - Bx₂)

4. Solve for 'a' and 'b': Once you have A and B, solve for 'a' and 'b':

   a = e^A b = e^B

5. Write the Equation: Substitute the values of 'a' and 'b' into y = ab^x.

Example: Using the same points (1, 6) and (3, 24):

1. Take the logarithm:

   ln(6) = ln(a) + ln(b) ln(24) = ln(a) + 3ln(b)

2. Let Y = ln(y), A = ln(a), B = ln(b):

   ln(6) = A + B ln(24) = A + 3B

3. Solve for B and A: Subtracting the first equation from the second:

   ln(24) - ln(6) = 2B B = ln(4) / 2 ≈ 0.693

   Substituting B back into ln(6) = A + B:

   A = ln(6) - ln(4) / 2 ≈ 0.896

4. Solve for 'a' and 'b':

   a = e^A ≈ e^0.896 ≈ 2.45 b = e^B ≈ e^0.693 ≈ 2

5. Write the equation:

   The equation is approximately y = 2.45(2)^x. Note that there might be a slight difference due to rounding off during calculations.

Handling Special Cases

• Point (0, a): If one of your points is (0, y₀), then 'a' is simply y₀ because y = ab⁰ = a. You only need to solve for 'b' using the second point.

• Negative 'b': While less common, an exponential function can have a negative base 'b' if the x-values are integers. However, the function will oscillate between positive and negative values. This requires careful interpretation and often doesn't represent realistic growth or decay models.

• Points yielding b=1: If your calculations lead to b=1, then you do not have an exponential function. A constant function (y = a) or a different type of function is needed.

Choosing the Best Method

Both methods are equally valid. The simultaneous equations method is generally simpler for calculations when dealing with only two points. However, the logarithmic transformation method is more robust and adaptable if you have multiple data points or want to apply statistical analysis to find the best-fitting exponential curve.

Importance of Accurate Calculations

Remember that minor inaccuracies in calculations can significantly affect the final equation. Use a calculator or software that offers sufficient precision, especially when dealing with logarithms and exponential functions. Rounding off should be done judiciously only at the final stage.

Applications of Exponential Equations

Exponential equations are crucial in various fields:

• Population Growth: Modeling the growth of populations (bacteria, animals, humans).
• Radioactive Decay: Describing the decay of radioactive substances.
• Compound Interest: Calculating interest earned on investments.
• Spread of Diseases: Analyzing the spread of infectious diseases.
• Cooling/Heating: Determining the rate of temperature change.

Mastering the ability to find the equation of an exponential graph from two points is a fundamental skill with far-reaching applications in mathematics and science. By understanding the methods outlined above and practicing with different examples, you can confidently tackle these problems and apply your knowledge to various real-world scenarios. Remember to always check your solutions by substituting your calculated equation back into the original points to ensure accuracy.