Question

d)
options:
$f(x) = -2^{x - 2}$
$f(x) = -2^{x}$
$f(x) = 2^{x - 2}$
$f(x) = 2^{x}$

Explanation:

Step1: Analyze the y - intercept

To find the y - intercept, we set \(x = 0\) in the function. Let's check each function:

• For \(f(x)=- 2^{x - 2}\), when \(x = 0\), \(f(0)=-2^{-2}=-\frac{1}{4}\), which is negative, but the graph passes through \((0,0)\) or near \((0,0)\) with non - negative value, so this is incorrect.
• For \(f(x)=-2^{x}\), when \(x = 0\), \(f(0)=-1\), negative, incorrect.
• For \(f(x)=2^{x-2}\), when \(x = 0\), \(f(0)=2^{-2}=\frac{1}{4}\approx0.25\).
• For \(f(x)=2^{x}\), when \(x = 0\), \(f(0)=1\).

The graph seems to pass through \((0,0)\) or very close to it. Let's also check the behavior as \(x\) increases. The function is an exponential growth function (since it is increasing rapidly for positive \(x\)).

Step2: Check the horizontal shift

The parent function \(y = 2^{x}\) has a y - intercept at \((0,1)\). The function \(y=2^{x - 2}\) is a horizontal shift of \(y = 2^{x}\) to the right by 2 units? Wait, no, the exponent rule: \(a^{x - h}\) is a shift of \(a^{x}\) by \(h\) units. Wait, let's re - evaluate the y - intercept. Wait, maybe we made a mistake. Let's check the value at \(x = 2\) for \(f(x)=2^{x-2}\): when \(x = 2\), \(f(2)=2^{0}=1\). For \(f(x)=2^{x}\), when \(x = 2\), \(f(2)=4\). Looking at the graph, when \(x = 2\), the value is around 1 (since at \(x = 2\), the graph is at \(y = 1\) approximately).

Wait, let's check the function \(f(x)=2^{x - 2}\). Let's rewrite \(2^{x-2}=\frac{2^{x}}{4}\). The graph of \(y = 2^{x}\) is a standard exponential growth curve. The graph of \(y=2^{x-2}\) is a vertical compression (by a factor of \(\frac{1}{4}\)) and horizontal shift? No, \(y = 2^{x-2}\) is a horizontal shift of \(y = 2^{x}\) to the right by 2 units? Wait, no, the formula \(y=a^{x - h}\) shifts the graph of \(y = a^{x}\) to the right by \(h\) units when \(h>0\). But when \(x = 0\), \(y=2^{-2}=\frac{1}{4}\), and when \(x = 2\), \(y = 2^{0}=1\), when \(x = 4\), \(y=2^{2}=4\). The graph in the picture, when \(x = 4\), the value is around 4? Wait, no, the graph at \(x = 4\) is around 4? Wait, the y - axis has marks at 2,4,6. At \(x = 4\), the graph is at \(y = 4\)? Wait, maybe my initial y - intercept analysis was wrong.

Wait, let's start over. The function is an exponential function of the form \(y = a^{x - h}\) or \(y=a^{x}\). The graph passes through \((0,0)\)? No, actually, when \(x = 0\), the graph is at \(y = 0\) (or very close). Wait, no, the graph is above the x - axis for all \(x\) (since it's an exponential function with positive base). Let's check the function \(f(x)=2^{x-2}\):

When \(x = 2\), \(f(2)=2^{0}=1\)

When \(x = 3\), \(f(3)=2^{1}=2\)

When \(x = 4\), \(f(4)=2^{2}=4\)

Which matches the graph's behavior: at \(x = 2\), \(y\approx1\); at \(x = 4\), \(y\approx4\). The function \(f(x)=2^{x}\) at \(x = 2\) is \(4\), which does not match. The functions with the negative sign are decreasing or have negative values, which do not match the graph (the graph is always non - negative and increasing). So the correct function is \(f(x)=2^{x - 2}\).

Answer:

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