Question

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Explanation:

Step1: Analyze the y-intercept

To find the y-intercept, set \( x = 0 \) in each function.

• For \( f(x)=-2^{x - 2} \): \( f(0)=-2^{-2}=-\frac{1}{4} \) (negative, doesn't match graph's y-intercept at \( y = 1 \) approximately).
• For \( f(x)=-2^{x} \): \( f(0)=-2^{0}=-1 \) (negative, doesn't match).
• For \( f(x)=2^{x - 2} \): \( f(0)=2^{-2}=\frac{1}{4} \) (too small, graph at \( x = 0 \) is around \( y = 1 \)? Wait, no, wait the graph: when \( x = 0 \), the graph is at \( y = 1 \)? Wait, let's check \( f(x)=2^{x} \): \( f(0)=2^{0}=1 \), which matches the y-intercept (the graph crosses the y-axis at \( y = 1 \)). Wait, but also check the shape: the graph is increasing, so the base is positive (since if it were negative, the function would be complex or have different behavior). Also, the horizontal asymptote: as \( x\to-\infty \), \( 2^{x}\to 0 \), which matches the left - hand side approaching the x - axis (y = 0). Let's check \( f(x)=2^{x - 2} \): at \( x = 2 \), \( f(2)=2^{0}=1 \), but the graph at \( x = 0 \) is 1? Wait no, wait the graph: when \( x = 0 \), the value is 1, when \( x = 1 \), \( 2^{1}=2 \), \( x = 2 \), \( 2^{2}=4 \), which matches the increasing curve. Wait, but let's re - evaluate:

Wait, the function \( f(x)=2^{x} \): when \( x = 0 \), \( y = 1 \); when \( x = 1 \), \( y = 2 \); when \( x = 2 \), \( y = 4 \), which matches the graph's shape (increasing, passing through (0,1), (1,2), (2,4)). The function \( f(x)=2^{x - 2} \) at \( x = 2 \) is 1, but the graph at \( x = 2 \) is at \( y = 4 \), so \( f(x)=2^{x} \) is correct.

Wait, let's check the horizontal asymptote: as \( x\to-\infty \), \( 2^{x}\to 0 \), which matches the left - hand side of the graph approaching the x - axis. The functions with negative signs are decreasing or have negative y - intercepts, so we can eliminate them. Between \( 2^{x} \) and \( 2^{x - 2} \): at \( x = 0 \), \( 2^{x}=1 \), \( 2^{x - 2}=\frac{1}{4} \). The graph at \( x = 0 \) is at \( y = 1 \), so \( f(x)=2^{x} \) is correct.

Step2: Check the shape and asymptote

The graph is an exponential growth curve (increasing, approaching y = 0 as \( x\to-\infty \)), which is the behavior of \( f(x)=a^{x} \) with \( a>1 \). The function \( f(x)=2^{x} \) has \( a = 2>1 \), so it's an exponential growth function, matching the graph's shape (increasing, horizontal asymptote at \( y = 0 \) as \( x\to-\infty \)).

Answer:

\( f(x)=2^{x} \) (the last option: \( f(x)=2^{x} \))