Question

which is the graph of $f(x) = \frac{1}{4}(4)^x$? images of four graphs with labeled points

Explanation:

Step1: Find the y-intercept (x=0)

Substitute \( x = 0 \) into \( f(x)=\frac{1}{4}(4)^x \).
\( f(0)=\frac{1}{4}(4)^0=\frac{1}{4}(1)=\frac{1}{4} \)? Wait, no—wait, \( 4^0 = 1 \), so \( \frac{1}{4}(1)=\frac{1}{4} \)? Wait, no, maybe I miscalculated. Wait, no, the function is \( f(x)=\frac{1}{4}(4)^x \). Let's recheck: \( 4^0 = 1 \), so \( f(0)=\frac{1}{4} \times 1=\frac{1}{4} \)? But the graphs have (0,4) or other points. Wait, maybe a typo? Wait, no, maybe the function is \( f(x)=\frac{1}{4}(4)^x \) or maybe \( f(x)=\frac{1}{4}(4)^x \) simplifies? Wait, \( \frac{1}{4}(4)^x = 4^{x - 1} \). Let's check \( x = 1 \): \( f(1)=\frac{1}{4}(4)^1=\frac{1}{4} \times 4 = 1 \). \( x = 2 \): \( f(2)=\frac{1}{4}(4)^2=\frac{1}{4} \times 16 = 4 \)? Wait, no, \( 4^2 = 16 \), \( \frac{1}{4} \times 16 = 4 \). Wait, but the third graph (rightmost) has a point (3,2) and (4,4). Wait, let's check the third graph: when \( x = 3 \), \( f(3)=\frac{1}{4}(4)^3=\frac{1}{4} \times 64 = 16 \)? No, that's not matching. Wait, maybe the function is \( f(x)=\frac{1}{4}(4)^x \) or maybe \( f(x)=\frac{1}{4}(4)^x \) is an exponential growth function because the base 4 > 1. So the graph should be increasing. Let's check the y-intercept: \( x = 0 \), \( f(0)=\frac{1}{4}(1)=\frac{1}{4} \), but the rightmost graph has a y-intercept near 0, then increases. Wait, the rightmost graph has a curve that starts near the x-axis, then rises. Let's check \( x = 2 \): \( f(2)=\frac{1}{4}(4)^2=\frac{1}{4} \times 16 = 4 \)? No, the rightmost graph has (4,4) and (3,2). Wait, maybe the function is \( f(x)=\frac{1}{4}(4)^x \) or maybe \( f(x)=4^{x - 1} \). Let's check \( x = 2 \): \( 4^{2 - 1}=4^1 = 4 \)? No, the rightmost graph at x=4 is (4,4), so when x=4, \( f(4)=\frac{1}{4}(4)^4=\frac{1}{4} \times 256 = 64 \), which doesn't match. Wait, maybe the function is \( f(x)=\frac{1}{4}(4)^x \) but the graphs are mislabeled? Wait, no, let's check the middle graph: (1,1). For \( f(1)=\frac{1}{4}(4)^1 = 1 \), which matches. \( x = 0 \): \( f(0)=\frac{1}{4}(4)^0=\frac{1}{4} \), but the middle graph has (0,4). Wait, that's a contradiction. Wait, maybe the function is \( f(x)=4(4)^{-x}=\frac{4}{4^x}=4^{1 - x} \), which is a decay function. Let's check \( x = 0 \): \( 4^{1 - 0}=4 \), so (0,4). \( x = 1 \): \( 4^{1 - 1}=1 \), so (1,1). \( x = 2 \): \( 4^{1 - 2}=\frac{1}{4} \), but the first graph has (2,1). Wait, the first graph: (0,4), (2,1). Let's check \( f(2) \) for \( f(x)=4(4)^{-x} \): \( 4(4)^{-2}=4 \times \frac{1}{16}=\frac{1}{4} \), no. Wait, the first graph has (2,1). Let's see: if \( f(2)=1 \), then \( \frac{1}{4}(4)^2 = 4 \), no. Wait, maybe the function is \( f(x)=\frac{1}{4}(4)^x \) is growth, and the rightmost graph is growth. Let's check the rightmost graph: when \( x = 3 \), the point is (3,2). Let's see \( f(3)=\frac{1}{4}(4)^3=\frac{1}{4} \times 64 = 16 \), no. Wait, maybe the function is \( f(x)=\frac{1}{4}(2)^x \)? No, the problem says \( \frac{1}{4}(4)^x \). Wait, maybe I made a mistake. Let's re-express \( f(x)=\frac{1}{4}(4)^x = 4^{x - 1} \). So it's an exponential function with base 4 (growth, since 4 > 1). So as \( x \) increases, \( f(x) \) increases. So the graph should be increasing, so the rightmost graph (the one with (3,2) and (4,4)) or the bottom one? Wait, the bottom one has (2,4). Let's check \( f(2)=4^{2 - 1}=4 \), so (2,4) would match \( f(2)=4 \). Wait, \( f(2)=\frac{1}{4}(4)^2=\frac{1}{4} \times 16 = 4 \), yes! So \( f(2)=4 \). So the bottom graph has (2,4). Let's check \( x = 1 \): \( f(1)=\frac{1}{4}(4)^1 = 1 \), no, the bottom graph doesn't show (1,1). Wait, the rightmost graph: (4,4), so \( f(4)=\frac{1}{4}(4)^4=\frac{1}{4} \times 256 = 64 \), no. Wait, I'm confused. Wait, let's list the points:

For \( f(x)=\frac{1}{4}(4)^x \):

- \( x = 0 \): \( \frac{1}{4}(1)=\frac{1}{4} \) (but graphs have (0,4) or near 0)
- \( x = 1 \): \( \frac{1}{4}(4)=1 \)
- \( x = 2 \): \( \frac{1}{4}(16)=4 \)
- \( x = 3 \): \( \frac{1}{4}(64)=16 \)

Wait, so at \( x = 2 \), \( f(2)=4 \), so the point (2,4) would be on the graph. The bottom graph has (2,4). Let's check the bottom graph: it's a growth curve, starting near the x-axis, then rising, with (2,4). That matches \( f(2)=4 \). Wait, but \( x = 0 \): \( f(0)=\frac{1}{4} \), so the y-intercept is \( \frac{1}{4} \), which is near the origin. The rightmost graph also has a low y-intercept. Wait, the bottom graph: let's see, when \( x = 2 \), \( f(2)=4 \), which is the point (2,4) on the bottom graph. So the bottom graph (the fourth one, with (2,4)) is the graph of \( f(x)=\frac{1}{4}(4)^x \), because at \( x = 2 \), \( f(2)=4 \), and it's an exponential growth curve (base 4 > 1, so increasing).

Step2: Analyze the function type

The function \( f(x)=\frac{1}{4}(4)^x = 4^{x - 1} \) is an exponential growth function (since the base \( 4 > 1 \)). So the graph should be increasing (as \( x \) increases, \( f(x) \) increases).

- The first two graphs are decay (decreasing), so eliminate them.
- The third graph: check \( x = 3 \), \( f(3)=\frac{1}{4}(4)^3 = 16 \), but the third graph has (3,2), which doesn't match.
- The bottom graph: \( x = 2 \), \( f(2)=4 \), which matches the point (2,4). It's a growth curve, so this is the correct graph.

Answer:

The graph with the point \((2, 4)\) (the bottom - most graph) is the graph of \( f(x)=\frac{1}{4}(4)^x \).