Problem 89 - Algebra and Trigonometry Real Mathematics, Real People

Question

Identify the domain, any intercepts, and any asymptotes of the function. \(f(x)=3^{x+1}+2\)

Step-by-Step Solution

Answer

The domain of the function \(f(x)=3^{x+1}+2\) is all real numbers or \(x \in R\). The function has a y-intercept at the point (0,5) and no x-intercept. The function has a horizontal asymptote at \(y=2\).

Step 1: Identify the Domain

In an exponential function such as \(f(x)=3^{x+1}+2\), the base (\(3\)) is raised to whatever value x takes. Since you can raise 3 to any power, either positive, negative, or zero, the domain of the function is all real numbers. Therefore, the domain is \(-\infty < x < \infty\) or \(x \in R\).

Step 2: Identify the Intercepts

To find the x-intercept, set the function equal to 0 and solve for x: \(0=3^{x+1}+2\) . This however doesn't yield any real solution so there's no x-intercept. To find the y-intercept, substitute \(x=0\) into the function to get \(y=3^{0+1}+2\), which is \(y=3+2=5\) . So the y-intercept is at the point (0,5).

Step 3: Identify the Asymptotes

The graph of an exponential function like \(f(x)=3^{x+1}+2\) has a horizontal asymptote, which is a horizontal line that the graph approaches as \(x\) approaches negative infinity. Subtracting 2 from the function, we get \(y-2=3^{x+1}\). As \(x\) approaches negative infinity, \(3^{x+1}\) approaches 0 because any non-zero number raised to an increasingly large negative power gets closer and closer to 0. That means \(y-2\) or \(y\) approaches 2. So the function has a horizontal asymptote at \(y=2\).

Key Concepts

Understanding Domain and Range of Exponential Functions

Exponential functions are intriguing, primarily because of their unique growth patterns and behaviors. The domain of an exponential function refers to all the possible x-values you can input into the function. For the function \(f(x) = 3^{x+1} + 2\), you can substitute any real number for \(x\). This is because it doesn't matter what power you raise the number \(3\) to—positive, negative, or even zero—you will always get a valid real number as a result. Hence, the domain for this function is all real numbers, expressible as \(-\infty < x < \infty\).

When we talk about the range, we're considering all possible output values that the function can produce, which are determined by how \(f(x)\) behaves. For exponential functions of this type, the output will never go below a certain value, in this case, \(2\) (due to the constant \(+2\) being added). The range of the function \(f(x) = 3^{x+1} + 2\) is therefore \(y > 2\). This means the lowest value that \(f(x)\) can approach is slightly above \(2\), stretching up to infinity.

Key points to remember:

• The domain of exponential functions is all real numbers.
• The range typically starts above a certain minimum value and extends to infinity.

Finding the Y-Intercept in Exponential Functions

The y-intercept of a function is the point where the graph intersects the y-axis. To find it, you simply need to determine the value of the function when \(x = 0\). This is because, at the y-axis, the x-coordinate is always zero.

For the function \(f(x) = 3^{x+1} + 2\), substitute 0 for \(x\) to find the y-intercept:

• \(f(0) = 3^{0+1} + 2\)
• \(f(0) = 3^1 + 2\)
• \(f(0) = 3 + 2 = 5\)

This tells us that the graph of the function will cross the y-axis at the point \((0, 5)\).

Remember:

• To find the y-intercept, set \(x = 0\) and solve for \(f(x)\).
• The y-intercept occurs at the point where the graph meets the y-axis.

Understanding Horizontal Asymptotes in Exponential Functions

A horizontal asymptote is a horizontal line that the graph of a function approaches but never quite reaches. In the context of exponential functions, this usually occurs because as \(x\) becomes very large or very small, the value of the function levels off at a certain horizontal line.

For \(f(x) = 3^{x+1} + 2\), we seek the value that \(y\) approaches as \(x\) goes to negative infinity. Examining the expression, we focus first on the part \(3^{x+1}\). As \(x\) decreases, \(3^{x+1}\) approaches zero, making the expression \(y - 2\) approach zero since we're subtracting \(2\). Thus, as \(x\) goes towards negative infinity, \(y\) gets closer and closer to \(2\). Therefore, the horizontal asymptote for this function is \(y = 2\).

Keep in mind:

• Horizontal asymptotes tell us about the end behavior of a function.
• They show where the function levels off as \(x\) becomes very large or very small.