Question

d)
graph with grid. options: ( f(x) = 5 - 2^x ), ( f(x) = 2^{1 - x} ) (selected), ( f(x) = 2^x - 5 ), ( f(x) = -2^{x - 1} )

Explanation:

Step1: Analyze the y-intercept

To find the y-intercept, set \( x = 0 \) in each function.
- For \( f(x)=5 - 2^{x} \): \( f(0)=5 - 2^{0}=5 - 1 = 4 \).
- For \( f(x)=2^{1 - x} \): \( f(0)=2^{1 - 0}=2^{1}=2 \)? Wait, no, wait. Wait, when \( x = 0 \), \( 2^{1-0}=2^1 = 2 \)? But the graph crosses the y-axis at \( y = 0 \) or near? Wait, looking at the graph, when \( x = 0 \), the y-value is 0? Wait, no, the graph at \( x = 0 \) is at \( y = 0 \)? Wait, no, the blue line at \( x = 0 \) is at \( y = 0 \)? Wait, maybe I miscalculated. Wait, let's check \( f(x)=2^{1 - x} \) at \( x = 0 \): \( 2^{1-0}=2 \). No, that's not matching. Wait, maybe another approach. Let's check the end behavior. The graph is decreasing, and as \( x \) increases, it goes down. Let's check the function \( f(x)=2^{1 - x} \). We can rewrite \( 2^{1 - x}=\frac{2^{1}}{2^{x}}=\frac{2}{2^{x}} \). As \( x \to \infty \), \( 2^{x} \to \infty \), so \( f(x) \to 0 \). As \( x \to -\infty \), \( 2^{x} \to 0 \), so \( f(x) \to \frac{2}{0^+} \to \infty \)? Wait, no, the graph as \( x \to -\infty \) is approaching a horizontal line (the left end is flat, approaching a horizontal asymptote). Wait, the graph in the picture, as \( x \) decreases (goes to the left), it's approaching a horizontal line (y=0? No, the left end is at y=0? Wait, the blue line on the left is at y=0? Wait, maybe I made a mistake. Wait, let's check the other functions. \( f(x)=5 - 2^{x} \): as \( x \to \infty \), \( 2^{x} \to \infty \), so \( f(x) \to -\infty \), which matches the graph (as \( x \) increases, it goes down to -infinity). At \( x = 0 \), \( f(0)=5 - 1 = 4 \), but the graph at \( x = 0 \) is at y=0? No, maybe the graph's y-intercept is at (0,0)? Wait, no, the blue line at \( x = 0 \) is at y=0? Wait, maybe the correct function is \( f(x)=2^{1 - x} \). Wait, let's re-express \( 2^{1 - x}=2\times2^{-x}=2\times(\frac{1}{2})^{x} \), which is an exponential decay function (since the base \( \frac{1}{2} \) is between 0 and 1). So as \( x \) increases, \( (\frac{1}{2})^{x} \) decreases, so \( f(x) \) decreases. As \( x \to \infty \), \( f(x) \to 0 \)? No, \( 2\times(\frac{1}{2})^{x} \to 0 \) as \( x \to \infty \), and as \( x \to -\infty \), \( (\frac{1}{2})^{x}=2^{-x} \to \infty \), so \( f(x) \to \infty \). But the graph on the left (as \( x \to -\infty \)) is approaching a horizontal line (y=0? No, the left end of the graph is flat, near y=0? Wait, the graph in the picture, the left end is at y=0 (the blue line on the left is at y=0). Wait, maybe the function \( f(x)=2^{1 - x} \) is correct. Wait, let's check the selected option: the radio button is on \( f(x)=2^{1 - x} \). Let's verify the end behavior. As \( x \to \infty \), \( 2^{1 - x} \to 0 \)? No, \( 2^{1 - x}=\frac{2}{2^{x}} \), so as \( x \to \infty \), \( 2^{x} \to \infty \), so \( f(x) \to 0 \). But the graph as \( x \) increases is going down to -infinity? Wait, no, the graph in the picture, as \( x \) increases (moves to the right), the blue line goes down, but the function \( f(x)=2^{1 - x} \) as \( x \) increases should approach 0 from above (since it's positive). But the graph is below the x-axis? Wait, no, the blue line is below the x-axis? Wait, the y-axis: the grid lines, the middle line is y=0? Wait, the graph is crossing the y-axis at (0,0)? No, the blue line at x=0 is at y=0? Wait, maybe the function is \( f(x)=2^{1 - x} \) is correct, as the selected option. Alternatively, maybe I made a mistake in the y-intercept. Let's check \( f(x)=2^{1 - x} \) at x=1: \( 2^{1 - 1}=2^{0}=1 \). At x=2: \( 2^{1 - 2}=2^{-1}=0.5 \). At x=3: \( 2^{1 - 3}=2^{-2}=0.25 \). So it's decreasing, approaching 0 as x increases, and as x decreases, it increases (since \( 2^{1 - x} \) when x is negative: x=-1, \( 2^{1 - (-1)}=2^{2}=4 \); x=-2, \( 2^{1 - (-2)}=2^{3}=8 \), so it goes to infinity as x approaches -infinity, which matches the left end of the graph (the left end is flat, approaching a horizontal asymptote, which would be y=0? No, as x approaches -infinity, \( 2^{1 - x} \) approaches infinity, so the left end should go up, but the graph's left end is flat at y=0? Wait, maybe the graph is actually \( f(x)=2^{1 - x} \) is correct, as the selected option. So the answer is the function \( f(x)=2^{1 - x} \).

Answer:

\( f(x) = 2^{1 - x} \) (the option with the radio button selected, \( f(x) = 2^{1 - x} \))