Draw the graph of $$ f(x)=\left(\frac{1}{5}\right)^{x+3}+2 $$

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To graph the function $$ f(x) = \left(\frac{1}{5}\right)^{x+3} + 2 $$, we can follow these steps:

1. **Identify the basic shape**: The term $$ \left(\frac{1}{5}\right)^{x+3} $$ is an exponential decay function because the base $$ \frac{1}{5} $$ is less than 1. The function will decrease as $$ x $$ increases.

2. **Determine the vertical shift**: The entire function is shifted up by 2 units due to the "+2" at the end. This means that the horizontal asymptote of the function will be at $$ y = 2 $$.

3. **Find key points**:
   - When $$ x = -3 $$:
     $$
     f(-3) = \left(\frac{1}{5}\right)^{0} + 2 = 1 + 2 = 3
     $$
   - When $$ x = -2 $$:
     $$
     f(-2) = \left(\frac{1}{5}\right)^{1} + 2 = \frac{1}{5} + 2 = 2.2
     $$
   - When $$ x = -1 $$:
     $$
     f(-1) = \left(\frac{1}{5}\right)^{2} + 2 = \frac{1}{25} + 2 = 2.04
     $$
   - When $$ x = 0 $$:
     $$
     f(0) = \left(\frac{1}{5}\right)^{3} + 2 = \frac{1}{125} + 2 \approx 2.008
     $$
   - When $$ x = 1 $$:
     $$
     f(1) = \left(\frac{1}{5}\right)^{4} + 2 = \frac{1}{625} + 2 \approx 2.0016
     $$

4. **Behavior as $$ x \to \infty $$**: As $$ x $$ increases, $$ \left(\frac{1}{5}\right)^{x+3} $$ approaches 0, so $$ f(x) $$ approaches 2.

5. **Behavior as $$ x \to -\infty $$**: As $$ x $$ decreases, $$ \left(\frac{1}{5}\right)^{x+3} $$ increases without bound, so $$ f(x) $$ approaches infinity.

6. **Plot the points**: Using the points calculated above, we can plot them on a graph.

7. **Draw the asymptote**: Draw a dashed line at $$ y = 2 $$ to represent the horizontal asymptote.

8. **Sketch the curve**: The curve will start from a high value when $$ x $$ is very negative, decrease towards the asymptote at $$ y = 2 $$, and never actually touch it.

Here is a rough sketch of the graph:

```
y
|
|          *
|         *
|        *
|       *
|      *
|     *
|    *
|   *
|  *
| *
|*_________________________ x
```

The curve approaches $$ y = 2 $$ as $$ x $$ increases and starts from a high value as $$ x $$ decreases.
The graph of $$ f(x)=\left(\frac{1}{5}\right)^{x+3}+2 $$ is an exponential decay curve shifted up by 2 units. It has a horizontal asymptote at $$ y = 2 $$ and decreases towards this line as $$ x $$ increases. The curve starts from a high value as $$ x $$ decreases and approaches infinity.

Draw the graph of

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