To understand the relationship between the graphs of [tex]\( y = 3^{-x} \)[/tex] and [tex]\( y = \left(\frac{1}{3}\right)^x \)[/tex], let's first rewrite [tex]\( y = 3^{-x} \)[/tex] in a more familiar form.
1. Recall the negative exponent rule:
[tex]\[
3^{-x} = \left(\frac{1}{3}\right)^x
\][/tex]
2. This shows us that the function [tex]\( y = 3^{-x} \)[/tex] is equivalent to [tex]\( y = \left(\frac{1}{3}\right)^x \)[/tex].
3. Next, let's consider the graph of [tex]\( y = \left(\frac{1}{3}\right)^x \)[/tex]:
[tex]\[
y = \left(\frac{1}{3}\right)^x
\][/tex]
This is an exponential decay function, where the base [tex]\( \frac{1}{3} \)[/tex] is a fraction between 0 and 1.
4. Now, let's recall the behavior of exponential functions. For [tex]\( y = 3^x \)[/tex], the graph is an exponential growth function. The graph of [tex]\( y = \left(\frac{1}{3}\right)^x \)[/tex] is reflected across the y-axis compared to [tex]\( y = 3^x \)[/tex]. This is because [tex]\( \left(\frac{1}{3}\right)^x \)[/tex] is the same as [tex]\( 3^{-x} \)[/tex]. In other words, the exponent's sign change corresponds to a reflection over the y-axis.
Therefore, the graphs of [tex]\( y = 3^{-x} \)[/tex] and [tex]\( y = \left(\frac{1}{3}\right)^x \)[/tex] are indeed reflections across the [tex]\( y \)[/tex]-axis.
Thus, the correct answer is:
The graphs are reflected across the [tex]\( y \)[/tex]-axis.
