Problem 41
Question
Graph the exponential function using transformations. State the \(y\) -intercept, two additional points, the domain, the range, and the horizontal asymptote. $$f(x)=e^{x+1}-4$$
Step-by-Step Solution
Verified Answer
The y-intercept is (0, -1.282), with additional points (1, 3.389) and (-1, -3). The domain is all real numbers, the range is (-4, ∞), and the horizontal asymptote is y = -4.
1Step 1: Identify the Base Function
The base function for transformations is the exponential function \(f(x) = e^x\). The graph of \(f(x) = e^x\) passes through the point \((0, 1)\) and has a horizontal asymptote at \(y = 0\).
2Step 2: Apply Horizontal Transformation
The expression \(e^{x+1}\) represents a horizontal shift to the left by 1 unit of the base function \(e^x\). To verify, note that if \( f(x) = e^x\), then \(f(x+1) = e^{x+1}\) will shift the graph leftwards.
3Step 3: Apply Vertical Transformation
The subtraction of 4, i.e., \(-4\) in \(f(x) = e^{x+1} - 4\), is a vertical shift downwards by 4 units. Thus, the entire graph of \(e^{x+1}\) is lowered by 4 units.
4Step 4: State the Transformed y-intercept
For the y-intercept, set \(x = 0\) and compute: \[ f(0) = e^{0+1} - 4 = e^1 - 4 \approx 2.718 - 4 = -1.282.\]Thus, the y-intercept is \((0, -1.282)\).
5Step 5: Determine Two Additional Points
Choose two other values of \(x\) to find additional points:1. For \(x=1\), \[ f(1) = e^{1+1} - 4 = e^2 - 4 \approx 7.389 - 4 = 3.389. \] Thus, the point is \((1, 3.389)\).2. For \(x=-1\), \[ f(-1) = e^{-1+1} - 4 = e^0 - 4 = 1 - 4 = -3. \] Thus, the point is \((-1, -3)\).
6Step 6: State the Domain and Range
The domain of an exponential function is all real numbers, or \((-f, f )\). The vertical shift affects the range, making it \((-4, f)\).
7Step 7: Identify the Horizontal Asymptote
The horizontal shift does not affect the asymptote. However, the vertical shift downward by 4 units moves the asymptote from \(y = 0\) to \(y = -4\).
Key Concepts
Graphing TransformationsDomain and RangeHorizontal AsymptoteExponential Functions
Graphing Transformations
Graphing transformations involves shifting, stretching, or reflecting a graph compared to its base function. Here, we work with the exponential function, which has unique properties that make transformations straightforward yet important.
- Horizontal shifts move the graph left or right.
- Vertical shifts move the graph up or down.
- Reflections flip the graph over a specific axis.
Domain and Range
Understanding the domain and range of a function helps us comprehend where the function is defined and the values it can take.
- The **domain** of exponential functions like \( f(x) = e^x \) is all real numbers, symbolized by \( (-\infty, \infty) \).
- The **range** is transformed by vertical shifts. For our function, the transformation shifted the graph downward. As a result, the range changes from \( (0, \infty) \) to \( (-4, \infty) \).
Horizontal Asymptote
An exponential function generally approaches a horizontal line, known as an asymptote, as it progresses towards extremes on the x-axis.
- For the base function \( f(x) = e^x \), the horizontal asymptote is at \( y = 0 \).
- The vertical transformation affects the horizontal asymptote.For our function, it shifts down 4 units, from \( y = 0 \) to \( y = -4 \).
Exponential Functions
Exponential functions, such as \( f(x) = e^x \), describe rapid growth or decay processes, frequently occurring in real-world applications.Key features of exponential functions include:
- **Base Value**: The constant \( e \) is approximately 2.718, a natural exponential base widely used in mathematics and other scientific applications.
- **Growth Pattern**: The graph of an exponential function steadily increases (for positive exponents) or decreases (for negative exponents).
- **Transformations**: They can undergo various transformations which alter their graph, including shifting and reflecting. This transforms the behavior while preserving the primary exponential nature.
Other exercises in this chapter
Problem 41
Solve the logarithmic equations exactly. $$\log _{3}(2 x+1)=4$$
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Write each expression as a single logarithm. $$\frac{1}{2} \log _{b} x+\frac{2}{3} \log _{b} y$$
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Evaluate the logarithms exactly (if possible). $$\log _{1 / 4} 4096$$
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Solve the logarithmic equations exactly. $$\log _{2}(3 x-1)=3$$
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