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.