Problem 21 - College Algebra with Corequisite Support

Question

For the following exercises, find the formula for an exponential function that passes through the two points given. (-2,6) and (3,1)

Step-by-Step Solution

Answer

The exponential function is \( f(x) = 6^{3/5} \cdot \left(\frac{1}{6}\right)^{x/5} \).

Step 1: Set Up the General Form

An exponential function can be generally expressed as \( f(x) = ab^x \), where \( a \) and \( b \) are constants. Our goal is to determine the values of \( a \) and \( b \) that allow the function to pass through the given points \((-2, 6)\) and \((3, 1)\).

Step 2: Create a System of Equations

Since the function passes through \((-2, 6)\), substitute \(x = -2\) and \(f(x) = 6\) into the exponential equation: \[ 6 = ab^{-2} \] Similarly, for the point \((3, 1)\), substitute \(x = 3\) and \(f(x) = 1\): \[ 1 = ab^3 \] This gives us a system of two equations:1. \( 6 = ab^{-2} \) 2. \( 1 = ab^3 \)

Step 3: Solve for \(b\)

To find \(b\), first divide the second equation by the first:\[ \frac{ab^3}{ab^{-2}} = \frac{1}{6} \] Simplifying gives us \( b^5 = \frac{1}{6} \). So, solve for \(b\) by taking the fifth root:\[ b = \left(\frac{1}{6}\right)^{1/5} \]

Step 4: Solve for \(a\)

Substitute the value of \(b\) back into one of the original equations to solve for \(a\). Use the equation \( 6 = ab^{-2} \):\[ 6 = a\left(\left(\frac{1}{6}\right)^{1/5}\right)^{-2} \] Simplify and solve for \(a\):\[ 6 = a \cdot \left(6^{2/5}\right) \] Thus, \( a = \frac{6}{6^{2/5}} = 6^{3/5} \).

Step 5: Write the Exponential Function

Now that we have \(a = 6^{3/5}\) and \(b = \left(\frac{1}{6}\right)^{1/5}\), the exponential function is:\[ f(x) = 6^{3/5} \cdot \left(\left(\frac{1}{6}\right)^{1/5}\right)^x \]

Key Concepts

Solving Systems of Equations

To solve systems of equations, you simultaneously work with multiple equations to find values of unknowns that satisfy all equations in the system. For this exercise, we have two equations derived from the exponential function passing through the given points:

• For point (-2, 6): \[ 6 = ab^{-2} \]
• And for point (3, 1): \[ 1 = ab^3 \]

Both equations have the same variables, \(a\) and \(b\), making them a system of equations that needs to be solved together.
Solving them involves expressing one variable in terms of the other, then using substitution or elimination methods. By dividing one equation by the other, you can eliminate one variable, providing a straightforward path to find the other.

Exponential Equations

An exponential equation is a mathematical expression where the variable appears in the exponent. They're usually structured like this: \[ ab^x \]where \(a\) and \(b\) are constants that influence the position and growth rate of the graph, respectively.
Exponential functions model real-world scenarios like population growth or radioactive decay. These equations have unique properties, making them pivotal in algebra. They often require logarithms or exponent rules for solutions.
In this example, you see exponential equations in action through \[ 6 = ab^{-2} \] and\[ 1 = ab^3 \]. These equations are key in calculating \(a\) and \(b\) so that the curve passes through specific points.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying equations using algebraic rules. This skill is crucial in solving equations like the ones in this exercise. It involves changing the form of expressions to isolate variables, combine like terms, or factor expressions.

• Our task is to isolate \(b\) by dividing one equation by another, emphasizing algebraic simplification\[ \frac{ab^3}{ab^{-2}} = \frac{1}{6} \]leads directly to \[ b^5 = \frac{1}{6} \].
• Raising both sides to the power of \(\frac{1}{5}\) finds \(b\):\[ b = \left(\frac{1}{6}\right)^{1/5} \].

Once \(b\) is found, the same techniques help solve for \(a\) using substitution and further simplifications.
Algebraic manipulation thus plays a vital role in finding solutions through understanding and effectively handling equations.