Problem 31 - Algebra and Trigonometry with Analytic Geometry

Question

Find an exponential function of the form \(f(x)=b a^{-x}+c\) that has the given horizontal asymptote and \(y\)-intercept and passes through point \(P\). \(y=32 ; \quad y\)-intercept \(212 ; \quad P(2,112)\)

Step-by-Step Solution

Answer

The exponential function is \(f(x) = 180 \left( \frac{3}{2} \right)^{-x} + 32\).

Step 1: Understanding the Horizontal Asymptote

We know that for an exponential function of the form \(f(x) = b a^{-x} + c\), the horizontal asymptote is given by \(y = c\). The problem specifies that the horizontal asymptote is \(y = 32\), which tells us that \(c = 32\).

Step 2: Using the y-intercept

The function has a \(y\)-intercept of 212, which means it passes through the point \((0, 212)\). When \(x = 0\), the equation becomes \(f(0) = b\cdot a^0 + c = b + c = 212\). Since we already found \(c = 32\), it follows that \(b + 32 = 212\). Solving for \(b\), we get \(b = 212 - 32 = 180\).

Step 3: Using the Point P(2,112) to Find a

The function also passes through the point \(P(2, 112)\). Substituting \(x = 2\) and \(f(x) = 112\) into the function, we have \(112 = 180a^{-2} + 32\). Simplify to find \(180a^{-2} = 112 - 32\), which gives \(180a^{-2} = 80\). Solving for \(a^{-2}\), we divide both sides by 180: \(a^{-2} = \frac{80}{180}\). Simplify this to \(a^{-2} = \frac{4}{9}\), and thus \(a^2 = \frac{9}{4}\). So, \(a = \frac{3}{2}\), because \(a\) must be positive.

Step 4: Expressing the Exponential Function

Now that we have the values for \(b\), \(a\), and \(c\), we can express the exponential function. Substituting these values into the general form \(f(x) = b a^{-x} + c\), we get \(f(x) = 180 \left( \frac{3}{2} \right)^{-x} + 32\).

Key Concepts

Horizontal Asymptote in Exponential Functions

Understanding the horizontal asymptote is crucial in studying exponential functions. A horizontal asymptote is a horizontal line that the graph of a function approaches as the input, or x-value, goes to infinity or negative infinity. For an exponential function in the form \(f(x) = b a^{-x} + c\), this asymptote is given by the constant \(c\). It tells us that as \(x\) becomes very large or very small, the value of \(f(x)\) approaches \(c\).

In our exercise, the horizontal asymptote is \(y = 32\); thus, \(c = 32\). This means that as \(x\) heads towards positive infinity, the function \(f(x)\) will get closer and closer to 32 without actually reaching it.

Horizontal asymptotes provide valuable information about the end behavior of the function and help us understand how the graph of the function behaves for extreme values of \(x\).

Finding the Y-intercept of an Exponential Function

The \(y\)-intercept is where the graph of a function crosses the y-axis. In mathematical terms, it's the value of \(f(x)\) when \(x = 0\). For an exponential function in the form \(f(x) = b a^{-x} + c\), the \(y\)-intercept can be found by substituting \(x = 0\). This simplifies the equation to \(f(0) = b + c\).

In the exercise, the \(y\)-intercept is given as 212. Therefore, substituting in our equation, we get \(212 = b + 32\). Solving for \(b\), we find \(b = 180\).

Knowing the \(y\)-intercept helps us quickly locate where the graph starts from on the y-axis, which is often a key step in plotting the function.

Understanding and Solving Exponential Equations

Exponential equations are equations in which the variable appears in the exponent. Solving these equations often requires logarithmic principles, but sometimes, like in our exercise, they can be simplified directly.

We used the additional point \(P(2, 112)\) given in our problem to solve for the base \(a\) of the exponential term. By substituting \(x = 2\) and \(f(x) = 112\) into the equation, \(112 = 180a^{-2} + 32\) simplifies to \(180a^{-2} = 80\). Solving further, \(a^{-2} = \frac{4}{9}\), then \(a^2 = \frac{9}{4}\), ultimately yields \(a = \frac{3}{2}\).

This process illustrates how additional points can be crucial in determining unknown parameters when solving exponential equations. It shows the direct relationship between functions and their graphical interpretations.

Function Evaluation

Function evaluation involves finding a function's output for a given input. For exponential functions like \(f(x) = b a^{-x} + c\), evaluating the function means substituting specific values of \(x\) to determine \(f(x)\). This process is used to verify that the function meets all given conditions, such as passing through specific points.

In the exercise, we have checked for correctness by evaluating points like the given \(y\)-intercept and point \(P\). This verifies that as we calculate \(f(0) = 212\) and \(f(2) = 112\), our derived equation fits the specific points provided.

Function evaluation is a practical tool not only for verifying but also for graphing corrections or determining other graph points when detailed transitions for mapping the function are required.