Here’s the solution to the given problem:

1. Consider the function $f(x) = 2\left(\frac{1}{3}\right)^{-x}$, find the following:

a. The parent function

The parent function for this is an exponential function of the form: $f(x) = b^x$ Here, $b = \frac{1}{3}$ (the base of the exponential part of the function). Thus, the parent function is: $f(x) = \left(\frac{1}{3}\right)^x$

b. The vertical intercept of $f(x)$, written as an ordered pair

To find the vertical intercept, evaluate $f(x)$ at $x = 0$: $f(0) = 2\left(\frac{1}{3}\right)^{-0}$ Since $\left(\frac{1}{3}\right)^0 = 1$: $f(0) = 2(1) = 2$ Thus, the vertical intercept is: $(0, 2)$

c. The equation of the horizontal asymptote

For exponential functions like $f(x) = 2\left(\frac{1}{3}\right)^{-x}$, as $x \to \infty$, the output approaches 0 but never reaches it. Therefore, the horizontal asymptote is: $y = 0$

d. Sketch $f(x)$

The function $f(x) = 2\left(\frac{1}{3}\right)^{-x}$ can be rewritten as: $f(x) = 2(3^x)$ Key features of the graph:

• The graph grows rapidly for $x > 0$ due to $3^x$.
• The graph decreases as $x < 0$, approaching 0 asymptotically.
• Vertical intercept: $(0, 2)$.
• Horizontal asymptote: $y = 0$.

Would you like me to provide a precise graph of $f(x)$? If so, let me know!

Let me know if you want further clarification on any part.

Related Questions:

1. What are the transformations applied to the parent function in $f(x)$?
2. How does the negative exponent in $\left(\frac{1}{3}\right)^{-x}$ affect the graph?
3. What is the domain and range of the function $f(x) = 2(3^x)$?
4. Can you derive $f'(x)$ to analyze the rate of growth for this function?
5. What happens to the graph of $f(x)$ if the coefficient "2" is changed to another constant?

Tip: Always identify key points like intercepts and asymptotes before sketching exponential graphs for accuracy!