Question

use transformations of the graph of ( f(x)=2^{x} ) to graph the given function. be sure to graph and give the equation of the asymptote. use the graph to determine the functions domain and range. if applicable, use a graphing utility to confirm your hand - drawn graphs.
( g(x)=2^{x - 4} )
graph ( g(x)=2^{x - 4} ) and its asymptote. use the graphing tool to graph the function as a solid curve and the asymptote as a dashed line.
the equation of the asymptote for ( g(x)=2^{x - 4} ) is ( square ). (type an equation.)

Explanation:

Step1: Recall transformation of exponential functions

For an exponential function of the form \( y = a^{x - h}+k \), the horizontal asymptote is \( y = k \). In the function \( g(x)=2^{x}-4 \), we can rewrite it as \( g(x)=2^{x}+(- 4) \), where \( k=-4 \)? Wait, no, wait. Wait, the general form for vertical shift: if we have \( f(x)=2^{x} \), and \( g(x)=2^{x}-4 \), this is a vertical shift down by 4 units. The parent function \( f(x) = 2^{x} \) has a horizontal asymptote at \( y = 0 \) (since as \( x ightarrow-\infty \), \( 2^{x} ightarrow0 \)). When we shift the graph of \( f(x) \) vertically down by 4 units, the horizontal asymptote also shifts down by 4 units.

Step2: Determine the asymptote

The parent function \( f(x)=2^{x} \) has horizontal asymptote \( y = 0 \). For the function \( g(x)=2^{x}-4 \), which is \( f(x)-4 \), the vertical transformation (subtracting 4) shifts the graph down 4 units. So the horizontal asymptote of \( g(x) \) is \( y=- 4 \)? Wait, no, wait. Wait, no, the function is \( g(x)=2^{x}-4 \), so it's a vertical shift. The horizontal asymptote of an exponential function \( a^{x}+k \) is \( y = k \). Here, \( a = 2 \), and the function is \( 2^{x}-4=2^{x}+(-4) \), so \( k=-4 \)? Wait, no, that's not right. Wait, no, the parent function \( f(x)=2^{x} \) has horizontal asymptote \( y = 0 \). When we do \( g(x)=f(x)-4 \), we shift the graph down by 4, so the horizontal asymptote also shifts down by 4. So the horizontal asymptote is \( y=-4 \)? Wait, but looking at the graph, the dashed line is at \( y = - 4 \)? Wait, no, the graph in the picture shows a dashed line at \( y = 0 \)? Wait, no, the user's graph: the dashed line is along \( y = 0 \)? Wait, no, the function is \( g(x)=2^{x}-4 \). Wait, maybe I made a mistake. Wait, no, let's re - express. The function \( g(x)=2^{x}-4 \): as \( x ightarrow-\infty \), \( 2^{x} ightarrow0 \), so \( g(x) ightarrow0 - 4=-4 \). So as \( x ightarrow-\infty \), \( g(x) ightarrow - 4 \), so the horizontal asymptote is \( y=-4 \)? But the graph in the picture: the dashed line is at \( y = 0 \)? Wait, maybe the function is \( g(x)=2^{x - 4}\)? Wait, the original problem says \( g(x)=2^{x}-4 \) or \( g(x)=2^{x - 4} \)? Wait, the user wrote \( g(x)=2^{x}-4 \). Wait, but the graph shown has a dashed line at \( y = 0 \)? Wait, no, looking at the graph, the dashed line is along \( y = 0 \)? Wait, maybe there's a typo, but according to the function \( g(x)=2^{x}-4 \), let's re - check.

Wait, the general rule: for \( y = a^{x}+k \), horizontal asymptote is \( y = k \). For \( y=a^{x - h}+k \), horizontal asymptote is \( y = k \). In the function \( g(x)=2^{x}-4 \), it's \( y = 2^{x}+(-4) \), so \( k=-4 \), so horizontal asymptote is \( y=-4 \). But maybe the function is \( g(x)=2^{x - 4} \)? Wait, the user's problem: "g(x)=2^x - 4" or "g(x)=2^(x - 4)"? The way it's written is \( g(x)=2^{x}-4 \). Let's confirm with the limit. As \( x ightarrow-\infty \), \( 2^{x} ightarrow0 \), so \( g(x)=2^{x}-4 ightarrow0 - 4=-4 \). So the horizontal asymptote is \( y=-4 \). Wait, but the graph in the picture: the dashed line is at \( y = 0 \)? Maybe the graph is for a different function, but according to the function \( g(x)=2^{x}-4 \), the asymptote is \( y=-4 \). Wait, no, maybe I misread the function. Let me check again. The problem says "g(x)=2^x - 4" or "g(x)=2^(x - 4)"? The user wrote "g(x)=2^x - 4". So according to the function \( g(x)=2^{x}-4 \), the horizontal asymptote is \( y=-4 \). But wait, maybe the function is \( g(x)=2^{x - 4} \). Let's check that. If \( g(x)=2^{x - 4} \), then it's a horizontal shift. The parent function \( f(x)=2^{x} \) has asymptote \( y = 0 \), and a horizontal shift doesn't change the horizontal asymptote, so asymptote would be \( y = 0 \). But the problem says "g(x)=2^x - 4". Wait, the user's hand - written function: "g(x)=2^x - 4" or "g(x)=2^(x - 4)"? The way it's typeset: "g(x)=2^x - 4" (since it's written as \( 2^{x}-4 \)). So let's go with that.

So, the parent function \( f(x)=2^{x} \) has horizontal asymptote \( y = 0 \). The function \( g(x)=f(x)-4 \) is a vertical shift down by 4 units. So the horizontal asymptote of \( g(x) \) is \( y=-4 \)? Wait, no, that can't be. Wait, no, the horizontal asymptote of an exponential function \( a^{x}+k \) is \( y = k \). Here, the function is \( 2^{x}-4=2^{x}+(-4) \), so \( k = - 4 \), so horizontal asymptote is \( y=-4 \). But let's check with the graph. The graph in the picture shows a curve that starts near the dashed line (which is at \( y = 0 \))? Wait, maybe the function is \( g(x)=2^{x - 4} \). Let's re - examine the problem statement. The user wrote: "g(x)=2^x - 4" or "g(x)=2^(x - 4)"? The original problem: "g(x)=2^x - 4" (the exponent is x, then minus 4). But maybe it's a typo, and it's \( 2^{x - 4} \). Wait, the graph shown has a curve that looks like \( 2^{x - 4} \), because when \( x = 4 \), \( 2^{4 - 4}=2^{0}=1 \), and the curve passes near (4,1). But if it's \( 2^{x}-4 \), when \( x = 0 \), \( g(0)=2^{0}-4=1 - 4=-3 \), which is below the x - axis. But the graph in the picture has the curve above the dashed line (which is at \( y = 0 \)). So maybe the function is \( g(x)=2^{x - 4} \). Let's assume that maybe there's a typo, and the function is \( g(x)=2^{x - 4} \). Then, the parent function \( f(x)=2^{x} \) has horizontal asymptote \( y = 0 \). A horizontal shift (replacing x with x - 4) doesn't change the horizontal asymptote, so the horizontal asymptote of \( g(x)=2^{x - 4} \) is \( y = 0 \). But the problem says "g(x)=2^x - 4". Wait, this is confusing. Wait, let's check the problem statement again: "Use transformations of the graph of f(x)=2^x to graph the given function. Be sure to graph and give the equation of the asymptote. Use the graph to determine the function's domain and range. If applicable, use a graphing utility to confirm your hand - drawn graphs. g(x)=2^x - 4".

Wait, maybe I made a mistake in the transformation. The function \( g(x)=2^{x}-4 \): as \( x ightarrow-\infty \), \( 2^{x} ightarrow0 \), so \( g(x) ightarrow0 - 4=-4 \). So the horizontal asymptote is \( y=-4 \). But the graph in the picture shows a dashed line at \( y = 0 \). This is a contradiction. So maybe the function is \( g(x)=2^{x - 4} \). Let's check the exponent. If the function is \( g(x)=2^{x - 4} \), then it's a horizontal shift right by 4 units. The parent function \( f(x)=2^{x} \) has horizontal asymptote \( y = 0 \), and a horizontal shift doesn't affect the horizontal asymptote, so the asymptote is \( y = 0 \). And the graph in the picture: when \( x = 4 \), \( 2^{4 - 4}=1 \), which matches the curve's position. So maybe there's a typo in the problem statement, and the function is \( g(x)=2^{x - 4} \). Assuming that, then:

The parent function is \( f(x)=2^{x} \), which has a horizontal asymptote at \( y = 0 \). The function \( g(x)=2^{x - 4} \) is a horizontal shift of \( f(x) \) to the right by 4 units. Horizontal shifts do not change the horizontal asymptote of an exponential function. So the horizontal asymptote of \( g(x) \) is the same as that of \( f(x) \), which is \( y = 0 \).

Wait, this is a critical point. Let's re - check the problem statement. The user wrote: "g(x)=2^x - 4" (exponent x, then minus 4) or "g(x)=2^(x - 4)" (exponent (x - 4))? The way it's typeset in the problem: "g(x)=2^x - 4" (the minus 4 is outside the exponent). But the graph shown has a curve that is a horizontal shift of \( 2^{x} \) to the right by 4 units. So maybe it's a typo, and the function is \( g(x)=2^{x - 4} \). Given that the graph is provided, and the curve is similar to \( 2^{x - 4} \) (since at \( x = 4 \), the value is 1, which is \( 2^{0}=1 \)), we'll proceed with the assumption that the function is \( g(x)=2^{x - 4} \) (maybe a typo in the problem, with the minus sign being a hyphen for the exponent).

So, for the function \( g(x)=2^{x - 4} \), which is a horizontal shift of \( f(x)=2^{x} \) to the right by 4 units. The horizontal asymptote of an exponential function \( a^{x + h}+k \) is determined by the vertical shift (k). Since there is no vertical shift (k = 0) in \( g(x)=2^{x - 4} \) (it's \( 2^{(x - 4)}+0 \)), the horizontal asymptote is \( y = 0 \).

Given that the graph in the picture has a dashed line at \( y = 0 \), we'll go with the function being \( g(x)=2^{x - 4} \) (assuming a typo in the problem statement, where the minus 4 is part of the exponent). Therefore, the horizontal asymptote is \( y = 0 \).

Answer:

\( y = 0 \)