Question

a) $k = -2$\
b) $k = 2$\
c) $k = 3$\
d) $k = 1$

Explanation:

Step1: Identify the function type

The graph appears to be an exponential function, likely of the form \( f(x) = 3^{x + k} \) (assuming a horizontal shift). From the graph, we can see that when \( x = -2 \), the function value seems to be \( 3^0 = 1 \) (since it's on the horizontal asymptote or crossing a key point). Wait, actually, looking at the graph, when \( x = -2 \), let's check the function. If the original function is \( 3^x \), shifting it left or right. Wait, the graph passes through \( x = -2 \) maybe? Wait, the grid has x-axis from -4 to 4. Let's assume the function is \( f(x) = 3^{x + k} \), and we can see that when \( x = -2 \), \( f(x) = 3^0 = 1 \)? No, maybe the function is \( 3^{x + k} \) and when \( x = -2 \), \( 3^{-2 + k} = 1 \)? Wait, no, 3^0 = 1, so -2 + k = 0 → k = 2? Wait, no, maybe the function is \( 3^{x + k} \) and we see that at x = -2, the function is at y=1 (since it's approaching the asymptote or has a point there). Wait, let's re-examine. If the function is \( 3^{x + k} \), and when x = -2, 3^{-2 + k} = 1. Since 3^0 = 1, then -2 + k = 0 → k = 2. Wait, but let's check the options. Option B is k=2.

Step2: Verify the shift

If k=2, then the function is \( 3^{x + 2} \), which is a horizontal shift of \( 3^x \) to the left by 2 units. Looking at the graph, it seems to be shifted left, so when x = -2, it's at 3^0 = 1, which matches. So k=2 is correct.

Answer:

B. \( k = 2 \)