Problem 30 - Algebra and Trigonometry with Analytic Geometry

Question

Find an exponential function of the form \(f(x)=b a^{x}\) that has the given \(y\)-intercept and passes through the point \(P\). y-intercept 6; \(\quad P\left(2, \frac{3}{32}\right)\)

Step-by-Step Solution

Answer

The exponential function is \( f(x) = 6 \cdot \left( \frac{1}{8} \right)^x \).

Step 1: Understand the Exponential Function

We are looking for a function of the form \( f(x) = b \cdot a^x \). The function has a \( y \)-intercept of 6, which means when \( x = 0 \), \( f(x) = 6 \). Additionally, the function should pass through the point \( P(2, \frac{3}{32}) \). We will use these pieces of information to find \( b \) and \( a \).

Step 2: Use the y-intercept to find b

Since the \( y \)-intercept is 6, when \( x = 0 \): \[ f(0) = b \cdot a^0 = b \cdot 1 = b = 6. \]So, \( b = 6 \).

Step 3: Substitute b and the point P into the function

Substitute \( b = 6 \) and the point \( P(2, \frac{3}{32}) \) into the equation \( f(2) = 6 \cdot a^2 = \frac{3}{32} \).

Step 4: Solve for a

Now solve the equation \[ 6a^2 = \frac{3}{32} \] Divide both sides by 6: \[ a^2 = \frac{3}{32} \cdot \frac{1}{6} = \frac{3}{192}. \]Simplify \( \frac{3}{192} \): \[ a^2 = \frac{1}{64}. \]Take the square root on both sides: \[ a = \frac{1}{8}. \]

Step 5: Write the Function

Now that we have both \( b = 6 \) and \( a = \frac{1}{8} \), we can write the final exponential function as \[ f(x) = 6 \cdot \left( \frac{1}{8} \right)^x. \]

Key Concepts

Understanding the Y-Intercept

In any mathematical function, the y-intercept is the point where the function crosses the y-axis. This means it's where the value of the function is determined when the input (usually "x") is zero. For exponential functions like \( f(x) = b \cdot a^x \), identifying the y-intercept can provide immediate information about the constant "b."

• For our function, a y-intercept of 6 tells us that when \( x = 0 \), the output, or \( f(x), \) is 6.
• So, \( f(0) = b \cdot a^0 = b \cdot 1 = b = 6 \).

This highlights that the y-intercept directly helps in determining the starting point of the function, and in this case, it leads to \( b = 6 \).

Understanding the y-intercept is crucial because it sets the stage for defining the rest of the function.

Function Passing Through a Point

When working with functions, determining if it passes through a specific point helps us find unknown values, such as the base of the exponential "a" in our case. Here, the exponential function is supposed to pass through the point \( P(2, \frac{3}{32}) \).

• Substituting the known values, you use the coordinates \( (x, y) = (2, \frac{3}{32}) \).
• Our function becomes \( f(2) = 6 \cdot a^2 = \frac{3}{32} \).

This equation is central in figuring out the value of "a." Post substituting, solving this correctly ensures that our exponential function indeed goes through the point \( P \). By understanding these principles, you ensure accuracy in defining functions that's not just theoretical but also fitting to conditions provided.

Solving Equations

Solving equations forms the backbone of finding unknowns in given scenarios. In this exercise, calculating the value of "a" is achieved by unraveling the equation derived from substituting point P.

• Starting with \( 6a^2 = \frac{3}{32} \), the main goal is isolating "a."
• To simplify, divide both sides by 6:
  \[ a^2 = \frac{3}{32} \cdot \frac{1}{6} = \frac{3}{192} \]
• The equation is now easier to work with as \( a^2 = \frac{1}{64} \).

By taking the square root of both sides, you find \( a = \frac{1}{8} \).

Solving such equations step-by-step ensures that each part is clear, reducing any mistakes along the way. This also highlights critical algebraic manipulation skills necessary for working with any exponential problem.

Simplifying Fractions

Simplifying fractions is a common step in algebra that helps make calculations more manageable and results more logical. In this exercise, simplification was key to acquiring a workable expression.

• After dividing \( \frac{3}{32} \) by 6, you reach \( \frac{3}{192} \).
• This fraction can definitely be simplified further to \( \frac{1}{64} \) for clarity and ease.

Why simplify? It reduces complexity and can often reveal hidden patterns within the numbers themselves, making further computation or patterns easier to spot.

Whether in solving exponential functions or any algebraic endeavor, simplifying fractions aligns computations in an orderly fashion and avoids unnecessary confusion.