Question

which graph represents a reflection of $f(x) = \frac{1}{3}(9)^x$ across the $x$-axis?

Explanation:

Step1: Recall reflection rule

To reflect a function \( f(x) \) across the \( x \)-axis, we use the transformation \( y=-f(x) \). So for \( f(x)=\frac{1}{3}(9)^{x} \), the reflected function is \( g(x)=-\frac{1}{3}(9)^{x} \).

Step2: Analyze original function

The original function \( f(x)=\frac{1}{3}(9)^{x} \) is an exponential growth function (since the base \( 9>1 \)) with a \( y \)-intercept at \( x = 0 \): \( f(0)=\frac{1}{3}(9)^{0}=\frac{1}{3}(1)=\frac{1}{3} \). It increases as \( x \) increases.

Step3: Analyze reflected function

The reflected function \( g(x)=-\frac{1}{3}(9)^{x} \) will be an exponential decay - like function (but actually a reflection of growth) with a \( y \)-intercept at \( x = 0 \): \( g(0)=-\frac{1}{3}(9)^{0}=-\frac{1}{3} \)? Wait, no, wait: Wait, \( f(0)=\frac{1}{3} \), so reflecting across \( x \)-axis, \( g(0)=-f(0)=-\frac{1}{3} \)? Wait, no, let's recalculate. Wait, \( 9^{0}=1 \), so \( f(0)=\frac{1}{3}\times1=\frac{1}{3} \). So the reflection across \( x \)-axis is \( g(x)=-f(x)=-\frac{1}{3}(9)^{x} \). So at \( x = 0 \), \( g(0)=-\frac{1}{3} \). As \( x \) increases, \( 9^{x} \) increases, so \( g(x) \) becomes more negative (since it's multiplied by - 1). As \( x \) decreases (goes to negative infinity), \( 9^{x}=\frac{1}{9^{\vert x\vert}} \) approaches 0, so \( g(x) \) approaches 0 from the negative side (since \( - \frac{1}{3}\times \) a small positive number is a small negative number).

Now let's analyze the graphs:

- The first graph: Let's check the \( y \)-intercept. At \( x = 0 \), the graph seems to pass near \( y=- 1 \)? Wait, no, let's look at the shape. The original function \( f(x)=\frac{1}{3}(9)^{x} \) has a \( y \)-intercept at \( \frac{1}{3}\approx0.333 \). The reflection should have a \( y \)-intercept at \( - \frac{1}{3}\approx - 0.333 \). Wait, maybe I made a mistake. Wait, no, let's re - evaluate the \( y \)-intercept of the original function. Wait, \( f(0)=\frac{1}{3}(9)^{0}=\frac{1}{3} \approx 0.333 \). So the reflection across \( x \)-axis will have a \( y \)-intercept at \( - 0.333 \). Now, let's look at the first graph: The curve is decreasing (as \( x \) increases, \( y \) decreases) and approaches 0 as \( x\to-\infty \) (from below the \( x \)-axis? Wait, no, the first graph: when \( x = 0 \), the curve is at \( y\approx - 1 \)? Wait, maybe my initial analysis of the \( y \)-intercept is wrong. Wait, let's check the original function again. Wait, \( f(x)=\frac{1}{3}(9)^{x} \). When \( x = 0 \), \( y=\frac{1}{3}\approx0.333 \). So the reflection is \( y =-\frac{1}{3}(9)^{x} \), so at \( x = 0 \), \( y =-\frac{1}{3}\approx - 0.333 \). Now, let's look at the first graph: The curve at \( x = 0 \) is near \( y=-1 \)? Wait, maybe the graphs are drawn with different scales. Alternatively, let's think about the behavior. The original function \( f(x)=\frac{1}{3}(9)^{x} \) is increasing, so the reflection should be decreasing (since it's a reflection across \( x \)-axis). The first graph: as \( x \) increases (from - 8 to 2), the curve is decreasing (going down as \( x \) moves to the right), which matches the behavior of \( g(x)=-\frac{1}{3}(9)^{x} \) (since as \( x \) increases, \( 9^{x} \) increases, so \( - \frac{1}{3}(9)^{x} \) decreases). The second graph is increasing, which is the original function (not the reflection). The third graph: let's see, at \( x = 0 \), it's at \( y = 0 \) or \( y=-1 \)? No, the third graph seems to have a \( y \)-intercept at \( y=-2 \) or so, and it's increasing as \( x \) decreases, which is not the case for the reflection.

Wait, maybe I made a mistake in the \( y \)-intercept calculation. Wait, \( f(0)=\frac{1}{3}(9)^{0}=\frac{1}{3} \), so the reflection across \( x \)-axis is \( g(x)=-f(x) \), so \( g(0)=-\frac{1}{3} \). So the \( y \)-intercept of the reflected graph is \( -\frac{1}{3} \). Now, looking at the first graph: when \( x = 0 \), the curve is at \( y\approx - 1 \)? No, maybe the graph's \( y \)-axis is scaled. Alternatively, let's think about the general shape. The original function is an exponential growth (increasing) function. The reflection across \( x \)-axis will be a function that is decreasing (since as \( x \) increases, the original function increases, so the reflection decreases) and approaches 0 as \( x\to-\infty \) (from below the \( x \)-axis) and goes to \( -\infty \) as \( x\to\infty \).

The first graph: as \( x \) increases (moves to the right), the curve goes down (decreases), and as \( x \) decreases (moves to the left), it approaches the \( x \)-axis from below? Wait, no, the first graph's left - hand side (as \( x\to-\infty \)) is approaching the \( x \)-axis from above? No, the first graph has a red arrow going to the left (as \( x\to-\infty \)) along the \( x \)-axis (y = 0) and then as \( x \) increases, it goes down. Wait, maybe the first graph is the reflection. The second graph is the original function (exponential growth, increasing, \( y \)-intercept at \( \frac{1}{3}\approx0.333 \), but the second graph's \( y \)-intercept is at \( y = 1 \)? Wait, maybe the scales are different. Let's re - express the original function: \( f(x)=\frac{1}{3}(9)^{x}=\frac{1}{3}(3^{2})^{x}=\frac{1}{3}(3)^{2x} \). At \( x = 0 \), \( f(0)=\frac{1}{3} \). At \( x = 1 \), \( f(1)=\frac{1}{3}(9)=3 \). At \( x=-1 \), \( f(-1)=\frac{1}{3}(\frac{1}{9})=\frac{1}{27}\approx0.037 \).

The reflection \( g(x)=-f(x)=-\frac{1}{3}(9)^{x} \). At \( x = 1 \), \( g(1)=- 3 \). At \( x=-1 \), \( g(-1)=-\frac{1}{27}\approx - 0.037 \).

Now, the first graph: when \( x = 1 \), \( y \) is negative (below the \( x \)-axis) and as \( x \) increases, \( y \) becomes more negative. When \( x=-1 \), \( y \) is near 0 (below the \( x \)-axis? No, the first graph at \( x = 0 \) is at \( y\approx - 1 \)? Wait, maybe the graph's \( y \)-axis is scaled. Alternatively, the key is that the reflection of an exponential growth function across the \( x \)-axis is a function that is decreasing (since the original is increasing) and has a \( y \)-intercept that is the negative of the original's \( y \)-intercept.

The original function \( f(x)=\frac{1}{3}(9)^{x} \) has a \( y \)-intercept of \( \frac{1}{3} \), so the reflection has a \( y \)-intercept of \( -\frac{1}{3} \). The first graph seems to have a \( y \)-intercept near \( y=-1 \), but maybe the scale is different. The second graph has a \( y \)-intercept near \( y = 1 \), which is close to the original's \( y \)-intercept ( \( \frac{1}{3}\approx0.333 \), but maybe the graph is stretched). The third graph has a \( y \)-intercept at \( y=-2 \) or so and is increasing as \( x \) decreases, which is not the case for the reflection.

So the first graph is the reflection of \( f(x)=\frac{1}{3}(9)^{x} \) across the \( x \)-axis.

Answer:

The first graph (the one with the curve starting near \( x = 0 \), \( y\approx - 1 \) and decreasing as \( x \) increases, approaching the \( x \)-axis as \( x\to-\infty \))