The problem requires graphing the exponential function:

$f(x) = -3 \left(\frac{3}{5}\right)^x$

Step 1: Analyze the function

1. Base: The base of the exponential function is $\frac{3}{5}$, which is less than 1, so this is a decaying exponential function.
2. Negative sign: The negative multiplier $-3$ means the function is reflected over the x-axis.
3. Asymptote: The horizontal asymptote is $y = 0$, since the function approaches 0 as $x \to \infty$.

Step 2: Plot points

We calculate a few points by substituting $x$ values into $f(x)$:

1. For $x = 0$: $f(0) = -3 \left(\frac{3}{5}\right)^0 = -3(1) = -3$ So, $(0, -3)$ is a point.

2. For $x = 1$: $f(1) = -3 \left(\frac{3}{5}\right)^1 = -3 \times \frac{3}{5} = -\frac{9}{5} \approx -1.8$ So, $(1, -1.8)$ is a point.

3. For $x = -1$: $f(-1) = -3 \left(\frac{3}{5}\right)^{-1} = -3 \times \frac{5}{3} = -5$ So, $(-1, -5)$ is a point.

4. For $x = 2$: $f(2) = -3 \left(\frac{3}{5}\right)^2 = -3 \times \frac{9}{25} = -\frac{27}{25} \approx -1.08$ So, $(2, -1.08)$ is a point.

5. For $x = -2$: $f(-2) = -3 \left(\frac{3}{5}\right)^{-2} = -3 \times \frac{25}{9} \approx -8.33$ So, $(-2, -8.33)$ is a point.

Step 3: Draw the graph

1. Plot the five points:
   • $(0, -3)$
   • $(1, -1.8)$
   • $(-1, -5)$
   • $(2, -1.08)$
   • $(-2, -8.33)$
2. Draw a smooth curve through the points, decaying towards the asymptote $y = 0$ as $x \to \infty$.
3. Indicate the horizontal asymptote at $y = 0$.

Would you like me to generate a visual graph for this? Let me know!

Relative questions:

1. What does the negative sign in front of the function indicate for the graph?
2. Why does the function decay rather than grow?
3. How does changing the base of the exponential function affect the graph?
4. What happens to the graph if the multiplier changes from $-3$ to a positive 3?
5. How would you determine the domain and range of this function?

Tip: Always calculate key points near $x = 0$ for clarity when graphing exponential functions.