Question

indicate which of the following four graphs is the correct graph of this function.
$f(x) = \frac{1}{4^{3-x}}$
answer
(graphs are shown: four coordinate planes with curves, axes labeled x and y from -8 to 8, grids, and curve shapes as described in the problem.)

Explanation:

Step1: Simplify the function

First, we can rewrite the function \( f(x)=\frac{1}{4^{3 - x}} \) using the property of exponents \( \frac{1}{a^{m - n}}=a^{n - m} \). So, \( f(x) = 4^{x - 3} \).

Step2: Analyze the function's behavior

- **Horizontal asymptote**: For an exponential function \( y = a^{x - h}+k \) (here \( k = 0 \), \( h = 3 \), \( a = 4>1 \)), as \( x ightarrow-\infty \), \( 4^{x - 3} ightarrow0 \) (since \( a^{-\infty} ightarrow0 \) for \( a > 1 \)). So, the horizontal asymptote is \( y = 0 \).
- **Y - intercept**: To find the y - intercept, set \( x = 0 \). Then \( f(0)=4^{-3}=\frac{1}{64}\approx0 \).
- **Increasing/Decreasing**: Since the base \( a = 4>1 \), the function \( y = 4^{x - 3} \) is an increasing function (exponential growth function).

Step3: Match with the graphs

- The first graph (top - left): It starts near the origin and increases, but let's check the y - intercept. If \( x = 0 \), \( f(0)=\frac{1}{64}\approx0 \), but when \( x = 3 \), \( f(3)=4^{0}=1 \). Wait, maybe we made a mistake. Wait, no, let's re - express the function again. Wait, \( f(x)=\frac{1}{4^{3 - x}}=4^{x - 3} \). Let's check the value at \( x = 3 \): \( f(3)=4^{0}=1 \). At \( x = 4 \), \( f(4)=4^{1}=4 \). At \( x = 2 \), \( f(2)=4^{-1}=\frac{1}{4} \).
- The third graph (bottom - left): It has a horizontal asymptote at \( y = 4 \)? No, because as \( x ightarrow-\infty \), \( 4^{x - 3} ightarrow0 \), not 4. So the third graph is out.
- The fourth graph (bottom - right): It passes through the origin (\( x = 0,y = 0 \)), but \( f(0)=\frac{1}{64} eq0 \), so it's out.
- The second graph (top - right): It is a decreasing function, but our function \( y = 4^{x - 3} \) is increasing, so it's out.
- Wait, wait, maybe I messed up the sign. Let's re - do the exponent: \( \frac{1}{4^{3 - x}}=4^{-(3 - x)}=4^{x - 3} \), correct. So the function is \( y = 4^{x - 3} \), which is an increasing exponential function with horizontal asymptote \( y = 0 \). Let's check the first graph (top - left): when \( x = 3 \), \( y = 1 \); when \( x = 4 \), \( y = 4 \); when \( x = 2 \), \( y=\frac{1}{4} \). The graph starts near the x - axis (since as \( x ightarrow-\infty \), \( y ightarrow0 \)) and increases. Wait, but the top - left graph: let's see the grid. Each square is 1 unit? The top - left graph: when \( x = 0 \), the y - value is near 0, and as \( x \) increases, it goes up. The bottom - left graph has a horizontal asymptote at \( y = 4 \), which is wrong. The bottom - right graph passes through (0,0), which is wrong. The top - right graph is decreasing, which is wrong. Wait, maybe I made a mistake in the sign. Wait, \( \frac{1}{4^{3 - x}}=4^{x - 3} \), which is an increasing function. The top - left graph is increasing, starts near the x - axis (since as \( x ightarrow-\infty \), \( y ightarrow0 \)) and as \( x \) increases, \( y \) increases. Let's check \( x = 3 \): \( y = 1 \), which is on the graph (since at \( x = 3 \), the y - value is 1, which is on the grid). So the correct graph is the top - left graph (the first graph). Wait, no, wait the bottom - left graph: no, the bottom - left graph has a horizontal line at \( y = 4 \) when \( x ightarrow-\infty \), which is incorrect. The top - left graph: when \( x = 0 \), \( y\approx0 \), when \( x = 3 \), \( y = 1 \), when \( x = 4 \), \( y = 4 \), which matches the function \( y = 4^{x - 3} \). So the correct graph is the top - left graph (the first graph among the four). Wait, but let's check again. Wait, the function \( f(x)=\frac{1}{4^{3 - x}} \) can also be written as \( f(x)=( \frac{1}{4}) ^{3 - x}=(\frac{1}{4})^{- (x - 3)}=4^{x - 3} \), which is an increasing exponential function. So the graph should be increasing, with horizontal asymptote \( y = 0 \), and passing through \( (3,1) \). The top - left graph shows this behavior: it starts near the x - axis (as \( x ightarrow-\infty \)) and increases as \( x \) increases.

Answer:

The top - left graph (the first graph in the 2x2 grid of graphs)