The given function is $f(x) = - \left(\frac{1}{3}\right)^x + 2$.

Let’s analyze the key features of this function:

1. Transformation:

   • The base function is $\left(\frac{1}{3}\right)^x$, which is an exponential decay function.
   • The negative sign in front of the exponential indicates a reflection over the x-axis.
   • The "+2" is a vertical shift upwards by 2 units.

2. Asymptote:

   • The horizontal asymptote of the graph is at $y = 2$, due to the vertical shift.

3. Intercept:

   • The y-intercept is found by evaluating $f(0)$: $f(0) = -\left(\frac{1}{3}\right)^0 + 2 = -1 + 2 = 1$ So the graph crosses the y-axis at $(0, 1)$.

4. End Behavior:

   • As $x \to \infty$, the exponential term $\left(\frac{1}{3}\right)^x \to 0$, so the function approaches the asymptote at $y = 2$.
   • As $x \to -\infty$, the term $\left(\frac{1}{3}\right)^x \to \infty$, but due to the negative sign, $f(x) \to -\infty$.

5. Decreasing Nature:

   • Since $\left(\frac{1}{3}\right)^x$ is an exponential decay function, and we have a reflection, the function is decreasing for all values of $x$.

Summary of Features:

• Asymptote: $y = 2$
• y-intercept: $(0, 1)$
• Decreasing for all $x$
• Domain: $(-\infty, \infty)$
• Range: $(-\infty, 2)$

Would you like further details or have any questions?

5 Related Questions:

1. How does changing the base of the exponential function affect its graph?
2. What happens if we remove the negative sign in front of the exponential?
3. How would shifting the graph vertically or horizontally change the features?
4. Can you describe how the reflection over the x-axis impacts the graph’s behavior?
5. How would the function change if the base was greater than 1, such as $\left(\frac{3}{2}\right)^x$?

Tip:

When graphing exponential functions, always pay attention to the horizontal asymptote and whether the graph is increasing or decreasing based on the base of the exponent.