Question

b)

options:

• $f(x) = -2^{x - 2}$
• $f(x) = -2^x$
• $f(x) = 2^{x - 2}$
• $f(x) = 2^x$

Explanation:

Step1: Analyze the sign of the function

The graph is below the x - axis for positive x (and near the x - axis for negative x), so the function should be negative. This eliminates the options \(f(x)=2^{x - 2}\) and \(f(x)=2^{x}\) since exponential functions with positive base and no negative sign will be positive for all real x.

Step2: Find a point on the graph

Let's check the y - intercept. The graph passes through (0, - 1)? Wait, no, let's check x = 2. For \(f(x)=-2^{x-2}\), when x = 2, \(f(2)=-2^{2 - 2}=-2^{0}=-1\). Wait, looking at the graph, when x = 2, the y - value is - 1? Wait, the graph at x = 2: let's see the grid. The graph at x = 2 is at y=-1? Wait, no, maybe x = 0. Let's check x = 0 for \(f(x)=-2^{x-2}\). \(f(0)=-2^{-2}=-\frac{1}{4}\), but the graph at x = 0 is at y = 0? Wait, no, the graph starts at the origin? Wait, maybe another approach. The general form of an exponential function \(y = a\cdot b^{x - h}+k\). The graph is a reflection (since it's decreasing and negative) of an exponential growth. The function \(y=-2^{x - 2}\): when x approaches \(-\infty\), \(2^{x - 2}\) approaches 0, so y approaches 0 from below (since the negative sign). When x increases, \(2^{x - 2}\) increases, so y becomes more negative, which matches the graph. Let's check the other negative option: \(f(x)=-2^{x}\). When x = 0, \(f(0)=-1\), but the graph at x = 0 seems to be at y = 0? Wait, no, the graph in the picture: at x = 0, the curve is at (0,0)? Wait, maybe my mistake. Wait, the graph is a curve that starts at the origin (0,0) and then decreases. Wait, let's check \(f(x)=-2^{x - 2}\) at x = 2: \(f(2)=-2^{0}=-1\), which matches the graph (at x = 2, y=-1). For \(f(x)=-2^{x}\), at x = 2, \(f(2)=-4\), which does not match. So the correct function is \(f(x)=-2^{x - 2}\).

Answer:

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