Use the graph of yequals
e Superscript x 
and transformations to sketch the exponential function f left parenthesis x right parenthesis equals e Superscript negative x Baseline plus 5
. 
Determine the domain and range.​ Also, determine the​ y-intercept, and find the equation of the horizontal asymptote.

To sketch the function \( f(x) = e^{-x} + 5 \) using the graph of \( y = e^x \) and transformations, follow these steps:

### 1. **Base Function: \( y = e^x \)**
   - The graph of \( y = e^x \) is an increasing exponential function that passes through the point \((0, 1)\) and has a horizontal asymptote at \( y = 0 \).

### 2. **Transformation: Horizontal Reflection**
   - The function \( f(x) = e^{-x} \) is a horizontal reflection of \( y = e^x \) across the y-axis. This means it will decrease as \( x \) increases, instead of increasing.

### 3. **Transformation: Vertical Shift**
   - The function \( f(x) = e^{-x} + 5 \) is obtained by shifting the graph of \( f(x) = e^{-x} \) upward by 5 units. This means that every point on the graph of \( y = e^{-x} \) is moved up by 5 units, and the horizontal asymptote is also moved up to \( y = 5 \).

### 4. **Sketching the Graph**
   - Start by sketching the graph of \( y = e^{-x} \), which is a decreasing exponential function that approaches 0 as \( x \) increases.
   - Then, shift this graph upward by 5 units. The new graph will approach \( y = 5 \) as \( x \) increases but will never touch it.

### 5. **Determine the Domain and Range**
   - **Domain**: The domain of \( f(x) = e^{-x} + 5 \) is all real numbers \( x \) since there are no restrictions on \( x \) for exponential functions.
     \[
     \text{Domain: } (-\infty, \infty)
     \]
   - **Range**: Since the graph is shifted upward by 5 units and the original graph of \( e^{-x} \) has a range of \( (0, \infty) \), the range of \( f(x) \) is \( (5, \infty) \).
     \[
     \text{Range: } (5, \infty)
     \]

### 6. **Determine the \( y \)-Intercept**
   - The \( y \)-intercept occurs where \( x = 0 \). Substituting \( x = 0 \) into the function:
     \[
     f(0) = e^{-0} + 5 = 1 + 5 = 6
     \]
   - So, the \( y \)-intercept is at \( (0, 6) \).

### 7. **Horizontal Asymptote**
   - The horizontal asymptote is the line that the graph approaches as \( x \) goes to \( \infty \) or \( -\infty \). Since the graph of \( f(x) = e^{-x} + 5 \) approaches 5 as \( x \) increases, the horizontal asymptote is:
     \[
     \text{Horizontal Asymptote: } y = 5
     \]

### Summary
- **Domain**: \( (-\infty, \infty) \)
- **Range**: \( (5, \infty) \)
- **\( y \)-Intercept**: \( (0, 6) \)
- **Horizontal Asymptote**: \( y = 5 \)

Would you like any further details or have any questions?

Here are five related questions for further practice:

1. How would the graph change if the function were \( f(x) = 2e^{-x} + 5 \)?
2. What is the impact of adding a negative constant to \( e^{-x} \), such as \( f(x) = e^{-x} - 3 \)?
3. How would the domain and range change if the function were \( f(x) = e^{-x} + 3 \)?
4. What would be the horizontal asymptote if the function were \( f(x) = e^{-x} + c \), where \( c \) is a constant?
5. How would you find the x-intercept of \( f(x) = e^{-x} + 5 \), if it exists?

**Tip**: When working with transformations, remember that horizontal shifts affect the input \( x \) directly, while vertical shifts affect the entire function's output.

Use the graph of y = e^x and transformations to sketch the exponential function f(x) = e^(-x) + 5. Determine the domain and range. Also, determine the y-intercept, and find the equation of the horizontal asymptote.

Learn how to sketch the exponential function f(x) = e^(-x) + 5 using graph transformations. This problem covers key concepts like domain and range, finding the y-intercept, and identifying the horizontal asymptote. Suitable for students in Grades 10-12.

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