Question

which is the graph of $f(x) = \frac{1}{4}(4)^x$? four graphs are shown, each with a coordinate grid and a curve; the first two curves are decreasing, the last two are increasing; points on the graphs are marked, e.g., (0,1/4)?, (2,1), (1,1), (4,4), (3,2), (2,4), (1,1)

Explanation:

Step1: Find the y-intercept

To find the y - intercept, we set \(x = 0\) in the function \(f(x)=\frac{1}{4}(4)^{x}\). Substituting \(x = 0\) into the function, we use the property \(a^{0}=1\) (where \(a = 4\) here). So \(f(0)=\frac{1}{4}(4)^{0}=\frac{1}{4}\times1=\frac{1}{4}\)? Wait, no, wait. Wait, \(4^{0}=1\), so \(\frac{1}{4}\times1=\frac{1}{4}\)? Wait, no, maybe I made a mistake. Wait, no, let's recalculate. Wait, the function is \(f(x)=\frac{1}{4}(4)^{x}\). When \(x = 0\), \(4^{0}=1\), so \(f(0)=\frac{1}{4}\times1=\frac{1}{4}\)? But looking at the graphs, maybe I miscalculated. Wait, no, wait, maybe the function is \(f(x)=\frac{1}{4}(4)^{x}\), let's check \(x = 1\): \(f(1)=\frac{1}{4}(4)^{1}=\frac{1}{4}\times4 = 1\). \(x=2\): \(f(2)=\frac{1}{4}(4)^{2}=\frac{1}{4}\times16 = 4\)? Wait, no, that can't be. Wait, maybe the function is \(f(x)=\frac{1}{4}(4)^{-x}\)? No, the original problem says \(f(x)=\frac{1}{4}(4)^{x}\). Wait, let's check the graphs. The bottom - left graph has a point \((1,1)\) and \((2,4)\)? Wait, no, the bottom - left graph: when \(x = 1\), \(y = 1\); \(x = 2\), \(y = 4\)? Wait, \(f(1)=\frac{1}{4}\times4=1\), \(f(2)=\frac{1}{4}\times16 = 4\). Let's check the y - intercept: when \(x = 0\), \(f(0)=\frac{1}{4}\times1=\frac{1}{4}\)? But the bottom - left graph at \(x = 0\), what's the y - value? Wait, the bottom - left graph has a point \((0, \frac{1}{4})\)? Wait, no, the bottom - left graph: the curve starts near the x - axis on the left, goes through \((1,1)\) and \((2,4)\). Let's check the function's behavior. The function \(f(x)=\frac{1}{4}(4)^{x}\) is an exponential growth function because the base \(4>1\) and the coefficient \(\frac{1}{4}\) is positive. So the graph should be increasing. Now let's check the points:

- For \(x = 0\): \(f(0)=\frac{1}{4}(4)^{0}=\frac{1}{4}\times1=\frac{1}{4}\)? No, wait, \(4^{0}=1\), so \(\frac{1}{4}\times1=\frac{1}{4}\). But the bottom - left graph: when \(x = 0\), the y - value is close to 0? Wait, no, maybe I misread the function. Wait, the problem says \(f(x)=\frac{1}{4}(4)^{x}\). Let's check \(x = 1\): \(f(1)=\frac{1}{4}\times4 = 1\), \(x = 2\): \(f(2)=\frac{1}{4}\times16 = 4\), \(x=3\): \(f(3)=\frac{1}{4}\times64 = 16\). Now let's look at the graphs:

The bottom - left graph (the fourth graph, the one at the bottom left) has a point \((1,1)\) and \((2,4)\) (wait, the point \((2,4)\)? Wait, the bottom - left graph: the curve goes through \((1,1)\) and \((2,4)\)? Wait, the point labeled \((2,4)\)? Wait, no, the bottom - left graph: when \(x = 1\), \(y = 1\); \(x = 2\), \(y = 4\). Let's check the y - intercept: when \(x = 0\), \(f(0)=\frac{1}{4}\), so the graph should pass through \((0,\frac{1}{4})\), which is close to the origin on the y - axis. The bottom - left graph (the fourth graph) has a curve that starts near the x - axis (as \(x\to-\infty\), \(4^{x}\to0\), so \(f(x)=\frac{1}{4}(4)^{x}\to0\)) and then increases, passing through \((1,1)\) and \((2,4)\). Let's check the other graphs:

- The first graph (top - left) is decreasing, so it's an exponential decay (base between 0 and 1), but our function has base 4>1, so it should be increasing. So eliminate the top - left and top - middle (second) graphs, which are decreasing.
- The third graph (top - right) has a point \((3,2)\) and \((4,4)\). Let's check \(f(3)=\frac{1}{4}(4)^{3}=\frac{1}{4}\times64 = 16\), which is not 2. So that's wrong.
- The fourth graph (bottom - left) has a point \((1,1)\): \(f(1)=\frac{1}{4}(4)^{1}=1\), correct. \((2,4)\): \(f(2)=\frac{1}{4}(4)^{2}=4\), correct. And as \(x\to-\infty\), \(f(x)\to0\), which matches the left - hand behavior (approaching the x - axis). And the y - intercept: when \(x = 0\), \(f(0)=\frac{1}{4}\), which is a small positive value, matching the graph's starting point near the x - axis at \(x = 0\).

Step2: Analyze the type of function (growth/decay)

The function \(f(x)=\frac{1}{4}(4)^{x}\) is of the form \(y = ab^{x}\), where \(a=\frac{1}{4}>0\) and \(b = 4>1\). So it is an exponential growth function (the graph should be increasing as \(x\) increases). The first two graphs (top - left and top - middle) are decreasing (exponential decay, base between 0 and 1), so we can eliminate them. The third graph (top - right) has points that don't match the function's values (e.g., \(f(3)=16\), but the graph has \(y = 2\) at \(x = 3\)). The fourth graph (bottom - left) has points \((1,1)\) ( \(f(1)=1\)) and \((2,4)\) ( \(f(2)=4\)) which match the function's values, and it is an increasing exponential curve, so it's the correct graph.

Answer:

The bottom - left graph (the fourth graph, the one with the curve passing through \((1,1)\) and \((2,4)\))