Question

match each exponential function to its graph. h(x) = \frac{1}{3}(4)^x, k(x) = -\frac{1}{3}(4)^x, u(x) = \frac{1}{3}\left(\frac{1}{4}\
ight)^x, n(x) = -\frac{1}{3}\left(\frac{1}{4}\
ight)^x

Explanation:

Step1: Analyze \( h(x)=\frac{1}{3}(4)^x \)

The base \( 4>1 \), so it's an exponential growth function. The coefficient \( \frac{1}{3}>0 \), so it passes through \( (0, \frac{1}{3}) \) (since \( h(0)=\frac{1}{3}(4)^0=\frac{1}{3} \)) and increases as \( x \) increases. This matches the left - most graph (the one with the curve rising from near the x - axis, passing through \( (0,\frac{1}{3}) \) approximately).

Step2: Analyze \( k(x)=-\frac{1}{3}(4)^x \)

The base \( 4 > 1 \), but the coefficient \(-\frac{1}{3}<0\). So it's a reflection of \( h(x) \) over the x - axis. When \( x = 0 \), \( k(0)=-\frac{1}{3}(4)^0=-\frac{1}{3} \). As \( x \) increases, \( 4^x \) increases, so \( k(x) \) decreases (goes more negative). As \( x \) decreases (goes to negative infinity), \( 4^x \) approaches 0, so \( k(x) \) approaches 0 from below the x - axis.

Step3: Analyze \( u(x)=\frac{1}{3}(\frac{1}{4})^x=\frac{1}{3}(4)^{-x} \)

The base \( \frac{1}{4}<1 \), so it's an exponential decay function. The coefficient \( \frac{1}{3}>0 \), so it passes through \( (0,\frac{1}{3}) \) (since \( u(0)=\frac{1}{3}(\frac{1}{4})^0=\frac{1}{3} \)) and decreases as \( x \) increases. This matches the right - most of the two graphs (the one with the curve falling as \( x \) increases, passing through \( (0,\frac{1}{3}) \) approximately).

Step4: Analyze \( n(x)=-\frac{1}{3}(\frac{1}{4})^x \)

The base \( \frac{1}{4}<1 \), and the coefficient \(-\frac{1}{3}<0\). It's a reflection of \( u(x) \) over the x - axis. When \( x = 0 \), \( n(0)=-\frac{1}{3}(\frac{1}{4})^0=-\frac{1}{3} \). As \( x \) increases, \( (\frac{1}{4})^x \) decreases, so \( n(x) \) (which is negative times a decreasing positive number) increases (approaches 0 from below the x - axis as \( x\to+\infty \) and goes to \( -\infty \) as \( x\to-\infty \)).

Answer:

• \( h(x)=\frac{1}{3}(4)^x \) matches the left - hand graph (the one with the rising exponential curve).
• \( k(x)=-\frac{1}{3}(4)^x \) would match a graph that is a reflection of \( h(x) \) over the x - axis (falling as \( x \) increases from a negative y - intercept).
• \( u(x)=\frac{1}{3}(\frac{1}{4})^x \) matches the right - hand graph (the one with the falling exponential curve).
• \( n(x)=-\frac{1}{3}(\frac{1}{4})^x \) would match a graph that is a reflection of \( u(x) \) over the x - axis (rising as \( x \) increases from a negative y - intercept).

If we assume we are matching \( h(x) \) first: \( h(x)=\frac{1}{3}(4)^x \) corresponds to the left graph (the graph on the left side of the two - graph set).