Question

match each graph below with the appropriate function.
a)

options: ( f(x) = 2^x - 5 ), ( f(x) = -2^{x - 2} ), ( f(x) = 2^{2 - x} ), ( f(x) = 5 - 2^x )

Explanation:

Step1: Analyze the y-intercept

To find the y-intercept, set \( x = 0 \) in each function.
- For \( f(x)=2^{x}-5 \): \( f(0)=2^{0}-5 = 1 - 5=-4 \)
- For \( f(x)=-2^{x - 2} \): \( f(0)=-2^{-2}=-\frac{1}{4} \)
- For \( f(x)=2^{2 - x} \): \( f(0)=2^{2-0}=4 \)
- For \( f(x)=5 - 2^{x} \): \( f(0)=5 - 2^{0}=5 - 1 = 4 \) Wait, no, wait. Wait, the graph has a y-intercept at \( y = 4 \)? Wait, looking at the graph, when \( x = 0 \), the graph is at \( y = 4 \)? Wait, no, the graph in the image: when \( x = 0 \), the y - value is 4? Wait, let's re - check. Wait, the graph is a decreasing curve, starting from the left (as \( x\) approaches negative infinity) high, and decreasing as \( x\) increases. Wait, when \( x = 0 \), the y - intercept: let's check the functions again.

Wait, for \( f(x)=2^{2 - x}\), rewrite it as \( f(x)=2^{2}\times2^{-x}=4\times(\frac{1}{2})^{x}\), which is an exponential decay function (since the base of the exponential with \( x\) in the exponent is \( \frac{1}{2}<1\)), so it decreases as \( x\) increases. When \( x = 0 \), \( f(0)=4\times1 = 4 \), which matches the y - intercept of the graph (the graph crosses the y - axis at \( y = 4\)).

For \( f(x)=5 - 2^{x}\), when \( x = 0 \), \( f(0)=5 - 1 = 4 \), but as \( x\) increases, \( 2^{x}\) increases, so \( f(x)=5 - 2^{x}\) will decrease, but when \( x\) is large positive, \( f(x)\) will approach \( -\infty \), while \( f(x)=2^{2 - x}=4\times(\frac{1}{2})^{x}\) will approach 0 as \( x\) approaches \( +\infty \), which matches the graph (the graph approaches 0 as \( x\) approaches \( +\infty \)).

For \( f(x)=2^{x}-5 \), as \( x\) increases, \( 2^{x}\) increases, so \( f(x)\) increases, which is an exponential growth (since the base \( 2>1\)) shifted down, so it's increasing, but our graph is decreasing, so we can eliminate \( f(x)=2^{x}-5 \).

For \( f(x)=-2^{x - 2}\), this is a negative exponential function. When \( x\) increases, \( 2^{x - 2}\) increases, so \( f(x)\) becomes more negative, so the graph would be decreasing but going to \( -\infty \) as \( x\) increases, which doesn't match our graph (our graph approaches 0 as \( x\) approaches \( +\infty \)).

Now, let's check the end - behavior. As \( x ightarrow+\infty \):
- For \( f(x)=2^{2 - x}=4\times(\frac{1}{2})^{x}\), as \( x ightarrow+\infty \), \( (\frac{1}{2})^{x} ightarrow0 \), so \( f(x) ightarrow0 \), which matches the graph (the graph approaches the x - axis as \( x ightarrow+\infty \)).
- For \( f(x)=5 - 2^{x}\), as \( x ightarrow+\infty \), \( 2^{x} ightarrow+\infty \), so \( f(x) ightarrow-\infty \), which does not match the graph (the graph approaches 0, not \( -\infty \)).

So the function \( f(x)=2^{2 - x}\) has the correct y - intercept (\( y = 4\)) and the correct end - behavior (approaches 0 as \( x ightarrow+\infty \)) and is a decreasing function (exponential decay), which matches the graph.

Step2: Analyze the end - behavior

- For \( f(x)=2^{2 - x}\), as \( x ightarrow+\infty \), \( 2^{2 - x}=4\times2^{-x} ightarrow0 \) (since \( 2^{-x}=\frac{1}{2^{x}}\) and \( 2^{x} ightarrow+\infty \) as \( x ightarrow+\infty \)), so the graph approaches the x - axis (y = 0) as \( x ightarrow+\infty \), which matches the given graph.
- For the other functions:
- \( f(x)=2^{x}-5 \) is an exponential growth function (since the base \( 2>1\)) shifted down, so it increases as \( x\) increases, which does not match the decreasing graph.
- \( f(x)=-2^{x - 2} \) is a negative exponential function, and as \( x ightarrow+\infty \), it approaches \( -\infty \), which does not match the graph approaching 0.
- \( f(x)=5 - 2^{x} \) as \( x ightarrow+\infty \) approaches \( -\infty \), which does not match the graph approaching 0.

Answer:

\( f(x)=2^{2 - x} \)