Question

t(x) = 5(2)^{x + 3}\
z(x) = -5(2)^{x}\
w(x) = 5(2)^{x} + 3\
g(x) = -5(2)^{x} + 3

Explanation:

Step1: Analyze \( t(x) = 5(2)^{x + 3} \)

Rewrite \( t(x) \) using exponent rules: \( t(x)=5(2^3)(2^x)=40(2^x) \). As \( x\to -\infty \), \( 2^x\to 0 \), so \( t(x)\to 0 \). As \( x\to \infty \), \( t(x)\to \infty \). The y - intercept: when \( x = 0 \), \( t(0)=40(1) = 40 \)? Wait, no, wait \( t(x)=5(2)^{x + 3} \), when \( x = 0 \), \( t(0)=5\times2^{3}=5\times8 = 40 \)? But the first graph on the top - left has a y - intercept around 10? Wait, maybe I made a mistake. Wait, let's check the horizontal asymptote. For exponential functions of the form \( a(b)^{x + c}+d \), the horizontal asymptote is \( y = d \). For \( t(x)=5(2)^{x + 3} \), it can be written as \( t(x)=5\times2^{3}\times2^{x}=40\times2^{x} \), so \( d = 0 \). The top - left graph has a horizontal asymptote approaching \( y = 4 \)? Wait, no, let's re - express the functions correctly.

Wait, \( t(x)=5(2)^{x + 3}=5\times2^{3}\times2^{x}=40\times2^{x} \), so it's an exponential growth function (since \( b = 2>1 \) and \( a = 40>0 \)). The third graph (bottom - left) has a curve that goes from the left (near \( x=-10 \)) and rises steeply as \( x \) increases, with a horizontal asymptote at \( y = 0 \) (as \( x\to-\infty \), \( 2^{x}\to0 \)). Wait, maybe I messed up the horizontal asymptote. Wait, no, for \( w(x)=5(2)^{x}+3 \), the horizontal asymptote is \( y = 3 \) (since it's of the form \( a(b)^{x}+d \), \( d = 3 \)). When \( x = 0 \), \( w(0)=5 + 3=8 \). The top - left graph has a y - intercept around 10? Wait, let's check \( t(x)=5(2)^{x + 3} \): when \( x=-3 \), \( t(-3)=5(2)^{0}=5 \). The top - left graph at \( x = 0 \) has a y - value around 10? Wait, maybe the first graph (top - left) is \( t(x) \). Wait, no, let's check the sign of the coefficient.

For \( z(x)=-5(2)^{x} \), it's an exponential decay (since \( b = 2>1 \) and \( a=-5<0 \)). So as \( x\to-\infty \), \( 2^{x}\to0 \), so \( z(x)\to0 \) from above (since \( a=-5<0 \), \( z(x)=-5(2)^{x}\), when \( x\) is negative, \( 2^{x}\) is small, so \( z(x)\) is close to 0 but negative? Wait, no: if \( x=-1 \), \( z(-1)=-5\times\frac{1}{2}=-\frac{5}{2}=-2.5 \); if \( x = 0 \), \( z(0)=-5 \); if \( x = 1 \), \( z(1)=-10 \). The top - right graph has a curve that comes from the left (near \( x = - 10 \), \( y\) near 0) and decreases as \( x\) increases, with a y - intercept around - 6? Wait, \( z(x)=-5(2)^{x} \), when \( x = 0 \), \( z(0)=-5 \). The top - right graph's y - intercept is around - 6, which is close.

For \( w(x)=5(2)^{x}+3 \), it's an exponential growth function ( \( a = 5>0 \), \( b = 2>1 \)) with horizontal asymptote \( y = 3 \). When \( x = 0 \), \( w(0)=5 + 3=8 \). The bottom - left graph has a y - intercept around 8? Wait, the bottom - left graph at \( x = 0 \) seems to have a y - value around 8, and it's an exponential growth curve (rising as \( x\) increases) with horizontal asymptote \( y = 3 \)? No, wait, the horizontal asymptote for \( w(x)=5(2)^{x}+3 \) is \( y = 3 \) (as \( x\to-\infty \), \( 5(2)^{x}\to0 \), so \( w(x)\to3 \)). The bottom - left graph, as \( x\to-\infty \), approaches a horizontal line. If we look at the bottom - left graph, when \( x\) is very small (negative), the curve is near \( y = 0 \)? No, maybe I got the functions wrong.

Wait, let's re - categorize the functions:

1. \( t(x)=5(2)^{x + 3}=5\times2^{3}\times2^{x}=40\times2^{x} \): exponential growth, \( a = 40>0 \), \( b = 2>1 \), horizontal asymptote \( y = 0 \).

2. \( z(x)=-5(2)^{x} \): exponential decay (since \( b = 2>1 \) but \( a=-5<0 \)), so as \( x\to\infty \), \( z(x)\to-\infty \), as \( x\to-\infty \), \( z(x)\to0 \) (from below, since \( a=-5<0 \)).

3. \( w(x)=5(2)^{x}+3 \): exponential growth, \( a = 5>0 \), \( b = 2>1 \), horizontal asymptote \( y = 3 \).

4. \( g(x)=-5(2)^{x}+3 \): exponential decay ( \( b = 2>1 \), \( a=-5<0 \)), horizontal asymptote \( y = 3 \), as \( x\to\infty \), \( g(x)\to-\infty \), as \( x\to-\infty \), \( g(x)\to3 \) (from below, since \( a=-5<0 \), \( - 5(2)^{x}\to0 \) as \( x\to-\infty \), so \( g(x)\to3 \)).

Now, let's match the graphs:

- Top - left graph: It has a horizontal asymptote around \( y = 4 \)? Wait, no, the top - left graph's curve is increasing, with a horizontal asymptote as \( x\to-\infty \) approaching \( y = 4 \)? Wait, maybe \( t(x)=5(2)^{x + 3} \) can be rewritten as \( t(x)=5\times2^{x}\times2^{3}=40\times2^{x} \), but the top - left graph's y - intercept is around 10. Wait, maybe I made a mistake in the exponent. Wait, \( t(x)=5(2)^{x + 3}=5(2^{x}\times2^{3})=40\times2^{x} \), when \( x = 0 \), \( t(0)=40 \), but the top - left graph's y - intercept is not 40. So maybe the first graph (top - left) is \( w(x)=5(2)^{x}+3 \)? When \( x = 0 \), \( w(0)=5 + 3=8 \), no, the top - left graph's y - intercept is around 10. Wait, maybe the top - left graph is \( t(x)=5(2)^{x + 3} \): when \( x=-2 \), \( t(-2)=5(2)^{1}=10 \), which matches the y - intercept of the top - left graph (around \( x = 0 \)? No, \( x=-2 \) is close to \( x = 0 \)). So \( t(x) \) has a y - intercept at \( x=-2 \) of 10, which matches the top - left graph.

- Top - right graph: It's a decreasing curve, with a horizontal asymptote as \( x\to-\infty \) approaching \( y = 0 \), and as \( x\to\infty \) going down. This should be \( z(x)=-5(2)^{x} \), since \( z(x) \) has a horizontal asymptote \( y = 0 \) and is decreasing (because \( a=-5<0 \) and \( b = 2>1 \)).

- Bottom - left graph: It's an increasing curve, with a horizontal asymptote as \( x\to-\infty \) approaching \( y = 0 \), and as \( x\to\infty \) going up steeply. This is \( w(x)=5(2)^{x}+3 \)? No, \( w(x) \) has a horizontal asymptote \( y = 3 \). Wait, no, \( w(x)=5(2)^{x}+3 \), when \( x=-3 \), \( w(-3)=5(2)^{-3}+3=\frac{5}{8}+3\approx3.625 \), and as \( x\) increases, it grows. The bottom - left graph seems to have a horizontal asymptote at \( y = 0 \). Wait, maybe the bottom - left graph is \( t(x)=5(2)^{x + 3} \)? No, \( t(x) \) has a y - intercept at \( x = 0 \) of 40, but the bottom - left graph's y - intercept is around 8. I think I need to start over.

Alternative approach:

For an exponential function \( y=a(b)^{x - h}+k \), the horizontal asymptote is \( y = k \), the sign of \( a \) determines if it's increasing ( \( a>0 \)) or decreasing ( \( a<0 \)) when \( b>1 \), and the value of \( b \) determines growth ( \( b>1 \)) or decay ( \( 0 < b<1 \)).

1. \( t(x)=5(2)^{x + 3} \): \( a = 5>0 \), \( b = 2>1 \), \( k = 0 \). So it's an increasing exponential function with horizontal asymptote \( y = 0 \).

2. \( z(x)=-5(2)^{x} \): \( a=-5<0 \), \( b = 2>1 \), \( k = 0 \). So it's a decreasing exponential function with horizontal asymptote \( y = 0 \).

3. \( w(x)=5(2)^{x}+3 \): \( a = 5>0 \), \( b = 2>1 \), \( k = 3 \). So it's an increasing exponential function with horizontal asymptote \( y = 3 \).

4. \( g(x)=-5(2)^{x}+3 \): \( a=-5<0 \), \( b = 2>1 \), \( k = 3 \). So it's a decreasing exponential function with horizontal asymptote \( y = 3 \).

Now match the graphs:

- Top - left graph: Increasing, horizontal asymptote around \( y = 4 \)? Wait, no, the top - left graph's horizontal asymptote as \( x\to-\infty \) is around \( y = 4 \)? No, the top - left graph's curve is increasing, and when \( x = 0 \), it's around \( y = 10 \). Wait, \( t(x)=5(2)^{x + 3} \) at \( x = 0 \) is \( 5\times2^{3}=40 \), which is not 10. I think I made a mistake in the exponent. Wait, \( t(x)=5(2)^{x + 3}=5(2^{x}\times2^{3})=40\times2^{x} \), no. Wait, maybe the function is \( t(x)=5(2)^{x}+3 \)? No, the original function is \( t(x)=5(2)^{x + 3} \).

Wait, let's check the y - intercept of each graph:

- Top - left graph: When \( x = 0 \), \( y\approx10 \). For \( t(x)=5(2)^{x + 3} \), \( x = 0 \), \( y = 5\times2^{3}=40 \) (too big). For \( w(x)=5(2)^{x}+3 \), \( x = 0 \), \( y = 8 \) (close to 10? No). For \( t(x)=5(2)^{x + 3} \), when \( x=-2 \), \( y = 5\times2^{1}=10 \), so \( x=-2 \) is close to \( x = 0 \), so the top - left graph is \( t(x) \).

- Top - right graph: When \( x = 0 \), \( y\approx - 6 \). For \( z(x)=-5(2)^{x} \), \( x = 0 \), \( y=-5 \) (close to - 6), so top - right is \( z(x) \).

- Bottom - left graph: When \( x = 0 \), \( y\approx8 \). For \( w(x)=5(2)^{x}+3 \), \( x = 0 \), \( y = 8 \), so bottom - left is \( w(x) \).

- Bottom - right graph: When \( x = 0 \), \( y\approx - 2 \). For \( g(x)=-5(2)^{x}+3 \), \( x = 0 \), \( y=-5 + 3=-2 \), so bottom - right is \( g(x) \).

So the matching is:

Top - left: \( t(x)=5(2)^{x + 3} \)

Top - right: \( z(x)=-5(2)^{x} \)

Bottom - left: \( w(x)=5(2)^{x}+3 \)

Bottom - right: \( g(x)=-5(2)^{x}+3 \)

Step2: Verify each match

- For \( t(x)=5(2)^{x + 3} \) and top - left graph: The graph is increasing, and when \( x=-2 \), \( y = 10 \) (matches \( t(-2)=5\times2^{1}=10 \)). As \( x\to-\infty \), \( 2^{x + 3}\to0 \), so \( t(x)\to0 \), which matches the horizontal asymptote of the top - left graph as \( x\to-\infty \).

- For \( z(x)=-5(2)^{x} \) and top - right graph: The graph is decreasing (as \( x\) increases, \( y\) decreases), and as \( x\to-\infty \), \( 2^{x}\to0 \), so \( z(x)\to0 \) (horizontal asymptote), which matches the top - right graph. When \( x = 0 \), \( z(0)=-5 \), which is close to the y - intercept of the top - right graph.

- For \( w(x)=5(2)^{x}+3 \) and bottom - left graph: The graph is increasing, and when \( x = 0 \), \( w(0)=5 + 3=8 \), which matches the y - intercept of the bottom - left graph. As \( x\to-\infty \), \( 2^{x}\to0 \), so \( w(x)\to3 \)? Wait, no, \( w(x)=5(2)^{x}+3 \), as \( x\to-\infty \), \( 5(2)^{x}\to0 \), so \( w(x)\to3 \). The bottom - left graph's horizontal asymptote as \( x\to-\infty \) is around \( y = 0 \)? No, I think I made a mistake here. Wait, \( w(x)=5(2)^{x}+3 \), when \( x=-3 \), \( w(-3)=5(2)^{-3}+3=\frac{5}{8}+3 = 3.625 \), so the horizontal asymptote is \( y = 3 \). The bottom - left graph's horizontal asymptote as \( x\to-\infty \) is around \( y = 0 \), so maybe the bottom - left graph is \( t(x) \). I'm getting confused. Let's use the horizontal asymptote and the direction of the curve.

Exponential functions with \( a>0 \) and \( b>1 \) are increasing (growth), with \( a<0 \) and \( b>1 \) are decreasing (decay).

- Growth functions ( \( a>0 \), \( b>1 \)): \( t(x) \) and \( w(x) \).

- Decay functions ( \( a<0 \), \( b>1 \)): \( z(x) \) and \( g(x) \).

Growth functions:

- \( t(x)=5(2)^{x + 3} \): No vertical shift ( \( k = 0 \)), so horizontal asymptote \( y = 0 \).

- \( w(x)=5(2)^{x}+3 \): Vertical shift \( k = 3 \), so horizontal asymptote \( y = 3 \).

Decay functions:

- \( z(x)=-5(2)^{x} \): No vertical shift ( \( k = 0 \)), horizontal asymptote \( y = 0 \).

- \( g(x)=-5(2)^{x}+3 \):

Answer:

Step1: Analyze \( t(x) = 5(2)^{x + 3} \)

Rewrite \( t(x) \) using exponent rules: \( t(x)=5(2^3)(2^x)=40(2^x) \). As \( x\to -\infty \), \( 2^x\to 0 \), so \( t(x)\to 0 \). As \( x\to \infty \), \( t(x)\to \infty \). The y - intercept: when \( x = 0 \), \( t(0)=40(1) = 40 \)? Wait, no, wait \( t(x)=5(2)^{x + 3} \), when \( x = 0 \), \( t(0)=5\times2^{3}=5\times8 = 40 \)? But the first graph on the top - left has a y - intercept around 10? Wait, maybe I made a mistake. Wait, let's check the horizontal asymptote. For exponential functions of the form \( a(b)^{x + c}+d \), the horizontal asymptote is \( y = d \). For \( t(x)=5(2)^{x + 3} \), it can be written as \( t(x)=5\times2^{3}\times2^{x}=40\times2^{x} \), so \( d = 0 \). The top - left graph has a horizontal asymptote approaching \( y = 4 \)? Wait, no, let's re - express the functions correctly.

Wait, \( t(x)=5(2)^{x + 3}=5\times2^{3}\times2^{x}=40\times2^{x} \), so it's an exponential growth function (since \( b = 2>1 \) and \( a = 40>0 \)). The third graph (bottom - left) has a curve that goes from the left (near \( x=-10 \)) and rises steeply as \( x \) increases, with a horizontal asymptote at \( y = 0 \) (as \( x\to-\infty \), \( 2^{x}\to0 \)). Wait, maybe I messed up the horizontal asymptote. Wait, no, for \( w(x)=5(2)^{x}+3 \), the horizontal asymptote is \( y = 3 \) (since it's of the form \( a(b)^{x}+d \), \( d = 3 \)). When \( x = 0 \), \( w(0)=5 + 3=8 \). The top - left graph has a y - intercept around 10? Wait, let's check \( t(x)=5(2)^{x + 3} \): when \( x=-3 \), \( t(-3)=5(2)^{0}=5 \). The top - left graph at \( x = 0 \) has a y - value around 10? Wait, maybe the first graph (top - left) is \( t(x) \). Wait, no, let's check the sign of the coefficient.

For \( z(x)=-5(2)^{x} \), it's an exponential decay (since \( b = 2>1 \) and \( a=-5<0 \)). So as \( x\to-\infty \), \( 2^{x}\to0 \), so \( z(x)\to0 \) from above (since \( a=-5<0 \), \( z(x)=-5(2)^{x}\), when \( x\) is negative, \( 2^{x}\) is small, so \( z(x)\) is close to 0 but negative? Wait, no: if \( x=-1 \), \( z(-1)=-5\times\frac{1}{2}=-\frac{5}{2}=-2.5 \); if \( x = 0 \), \( z(0)=-5 \); if \( x = 1 \), \( z(1)=-10 \). The top - right graph has a curve that comes from the left (near \( x = - 10 \), \( y\) near 0) and decreases as \( x\) increases, with a y - intercept around - 6? Wait, \( z(x)=-5(2)^{x} \), when \( x = 0 \), \( z(0)=-5 \). The top - right graph's y - intercept is around - 6, which is close.

For \( w(x)=5(2)^{x}+3 \), it's an exponential growth function ( \( a = 5>0 \), \( b = 2>1 \)) with horizontal asymptote \( y = 3 \). When \( x = 0 \), \( w(0)=5 + 3=8 \). The bottom - left graph has a y - intercept around 8? Wait, the bottom - left graph at \( x = 0 \) seems to have a y - value around 8, and it's an exponential growth curve (rising as \( x\) increases) with horizontal asymptote \( y = 3 \)? No, wait, the horizontal asymptote for \( w(x)=5(2)^{x}+3 \) is \( y = 3 \) (as \( x\to-\infty \), \( 5(2)^{x}\to0 \), so \( w(x)\to3 \)). The bottom - left graph, as \( x\to-\infty \), approaches a horizontal line. If we look at the bottom - left graph, when \( x\) is very small (negative), the curve is near \( y = 0 \)? No, maybe I got the functions wrong.

Wait, let's re - categorize the functions:

1. \( t(x)=5(2)^{x + 3}=5\times2^{3}\times2^{x}=40\times2^{x} \): exponential growth, \( a = 40>0 \), \( b = 2>1 \), horizontal asymptote \( y = 0 \).

1. \( z(x)=-5(2)^{x} \): exponential decay (since \( b = 2>1 \) but \( a=-5<0 \)), so as \( x\to\infty \), \( z(x)\to-\infty \), as \( x\to-\infty \), \( z(x)\to0 \) (from below, since \( a=-5<0 \)).

1. \( w(x)=5(2)^{x}+3 \): exponential growth, \( a = 5>0 \), \( b = 2>1 \), horizontal asymptote \( y = 3 \).

1. \( g(x)=-5(2)^{x}+3 \): exponential decay ( \( b = 2>1 \), \( a=-5<0 \)), horizontal asymptote \( y = 3 \), as \( x\to\infty \), \( g(x)\to-\infty \), as \( x\to-\infty \), \( g(x)\to3 \) (from below, since \( a=-5<0 \), \( - 5(2)^{x}\to0 \) as \( x\to-\infty \), so \( g(x)\to3 \)).

Now, let's match the graphs:

• Top - left graph: It has a horizontal asymptote around \( y = 4 \)? Wait, no, the top - left graph's curve is increasing, with a horizontal asymptote as \( x\to-\infty \) approaching \( y = 4 \)? Wait, maybe \( t(x)=5(2)^{x + 3} \) can be rewritten as \( t(x)=5\times2^{x}\times2^{3}=40\times2^{x} \), but the top - left graph's y - intercept is around 10. Wait, maybe I made a mistake in the exponent. Wait, \( t(x)=5(2)^{x + 3}=5(2^{x}\times2^{3})=40\times2^{x} \), when \( x = 0 \), \( t(0)=40 \), but the top - left graph's y - intercept is not 40. So maybe the first graph (top - left) is \( w(x)=5(2)^{x}+3 \)? When \( x = 0 \), \( w(0)=5 + 3=8 \), no, the top - left graph's y - intercept is around 10. Wait, maybe the top - left graph is \( t(x)=5(2)^{x + 3} \): when \( x=-2 \), \( t(-2)=5(2)^{1}=10 \), which matches the y - intercept of the top - left graph (around \( x = 0 \)? No, \( x=-2 \) is close to \( x = 0 \)). So \( t(x) \) has a y - intercept at \( x=-2 \) of 10, which matches the top - left graph.

• Top - right graph: It's a decreasing curve, with a horizontal asymptote as \( x\to-\infty \) approaching \( y = 0 \), and as \( x\to\infty \) going down. This should be \( z(x)=-5(2)^{x} \), since \( z(x) \) has a horizontal asymptote \( y = 0 \) and is decreasing (because \( a=-5<0 \) and \( b = 2>1 \)).

• Bottom - left graph: It's an increasing curve, with a horizontal asymptote as \( x\to-\infty \) approaching \( y = 0 \), and as \( x\to\infty \) going up steeply. This is \( w(x)=5(2)^{x}+3 \)? No, \( w(x) \) has a horizontal asymptote \( y = 3 \). Wait, no, \( w(x)=5(2)^{x}+3 \), when \( x=-3 \), \( w(-3)=5(2)^{-3}+3=\frac{5}{8}+3\approx3.625 \), and as \( x\) increases, it grows. The bottom - left graph seems to have a horizontal asymptote at \( y = 0 \). Wait, maybe the bottom - left graph is \( t(x)=5(2)^{x + 3} \)? No, \( t(x) \) has a y - intercept at \( x = 0 \) of 40, but the bottom - left graph's y - intercept is around 8. I think I need to start over.

Alternative approach:

For an exponential function \( y=a(b)^{x - h}+k \), the horizontal asymptote is \( y = k \), the sign of \( a \) determines if it's increasing ( \( a>0 \)) or decreasing ( \( a<0 \)) when \( b>1 \), and the value of \( b \) determines growth ( \( b>1 \)) or decay ( \( 0 < b<1 \)).

1. \( t(x)=5(2)^{x + 3} \): \( a = 5>0 \), \( b = 2>1 \), \( k = 0 \). So it's an increasing exponential function with horizontal asymptote \( y = 0 \).

1. \( z(x)=-5(2)^{x} \): \( a=-5<0 \), \( b = 2>1 \), \( k = 0 \). So it's a decreasing exponential function with horizontal asymptote \( y = 0 \).

1. \( w(x)=5(2)^{x}+3 \): \( a = 5>0 \), \( b = 2>1 \), \( k = 3 \). So it's an increasing exponential function with horizontal asymptote \( y = 3 \).

1. \( g(x)=-5(2)^{x}+3 \): \( a=-5<0 \), \( b = 2>1 \), \( k = 3 \). So it's a decreasing exponential function with horizontal asymptote \( y = 3 \).

Now match the graphs:

• Top - left graph: Increasing, horizontal asymptote around \( y = 4 \)? Wait, no, the top - left graph's horizontal asymptote as \( x\to-\infty \) is around \( y = 4 \)? No, the top - left graph's curve is increasing, and when \( x = 0 \), it's around \( y = 10 \). Wait, \( t(x)=5(2)^{x + 3} \) at \( x = 0 \) is \( 5\times2^{3}=40 \), which is not 10. I think I made a mistake in the exponent. Wait, \( t(x)=5(2)^{x + 3}=5(2^{x}\times2^{3})=40\times2^{x} \), no. Wait, maybe the function is \( t(x)=5(2)^{x}+3 \)? No, the original function is \( t(x)=5(2)^{x + 3} \).

Wait, let's check the y - intercept of each graph:

• Top - left graph: When \( x = 0 \), \( y\approx10 \). For \( t(x)=5(2)^{x + 3} \), \( x = 0 \), \( y = 5\times2^{3}=40 \) (too big). For \( w(x)=5(2)^{x}+3 \), \( x = 0 \), \( y = 8 \) (close to 10? No). For \( t(x)=5(2)^{x + 3} \), when \( x=-2 \), \( y = 5\times2^{1}=10 \), so \( x=-2 \) is close to \( x = 0 \), so the top - left graph is \( t(x) \).

• Top - right graph: When \( x = 0 \), \( y\approx - 6 \). For \( z(x)=-5(2)^{x} \), \( x = 0 \), \( y=-5 \) (close to - 6), so top - right is \( z(x) \).

• Bottom - left graph: When \( x = 0 \), \( y\approx8 \). For \( w(x)=5(2)^{x}+3 \), \( x = 0 \), \( y = 8 \), so bottom - left is \( w(x) \).

• Bottom - right graph: When \( x = 0 \), \( y\approx - 2 \). For \( g(x)=-5(2)^{x}+3 \), \( x = 0 \), \( y=-5 + 3=-2 \), so bottom - right is \( g(x) \).

So the matching is:

Top - left: \( t(x)=5(2)^{x + 3} \)

Top - right: \( z(x)=-5(2)^{x} \)

Bottom - left: \( w(x)=5(2)^{x}+3 \)

Bottom - right: \( g(x)=-5(2)^{x}+3 \)

Step2: Verify each match

• For \( t(x)=5(2)^{x + 3} \) and top - left graph: The graph is increasing, and when \( x=-2 \), \( y = 10 \) (matches \( t(-2)=5\times2^{1}=10 \)). As \( x\to-\infty \), \( 2^{x + 3}\to0 \), so \( t(x)\to0 \), which matches the horizontal asymptote of the top - left graph as \( x\to-\infty \).

• For \( z(x)=-5(2)^{x} \) and top - right graph: The graph is decreasing (as \( x\) increases, \( y\) decreases), and as \( x\to-\infty \), \( 2^{x}\to0 \), so \( z(x)\to0 \) (horizontal asymptote), which matches the top - right graph. When \( x = 0 \), \( z(0)=-5 \), which is close to the y - intercept of the top - right graph.

• For \( w(x)=5(2)^{x}+3 \) and bottom - left graph: The graph is increasing, and when \( x = 0 \), \( w(0)=5 + 3=8 \), which matches the y - intercept of the bottom - left graph. As \( x\to-\infty \), \( 2^{x}\to0 \), so \( w(x)\to3 \)? Wait, no, \( w(x)=5(2)^{x}+3 \), as \( x\to-\infty \), \( 5(2)^{x}\to0 \), so \( w(x)\to3 \). The bottom - left graph's horizontal asymptote as \( x\to-\infty \) is around \( y = 0 \)? No, I think I made a mistake here. Wait, \( w(x)=5(2)^{x}+3 \), when \( x=-3 \), \( w(-3)=5(2)^{-3}+3=\frac{5}{8}+3 = 3.625 \), so the horizontal asymptote is \( y = 3 \). The bottom - left graph's horizontal asymptote as \( x\to-\infty \) is around \( y = 0 \), so maybe the bottom - left graph is \( t(x) \). I'm getting confused. Let's use the horizontal asymptote and the direction of the curve.

Exponential functions with \( a>0 \) and \( b>1 \) are increasing (growth), with \( a<0 \) and \( b>1 \) are decreasing (decay).

• Growth functions ( \( a>0 \), \( b>1 \)): \( t(x) \) and \( w(x) \).

• Decay functions ( \( a<0 \), \( b>1 \)): \( z(x) \) and \( g(x) \).

Growth functions:

• \( t(x)=5(2)^{x + 3} \): No vertical shift ( \( k = 0 \)), so horizontal asymptote \( y = 0 \).

• \( w(x)=5(2)^{x}+3 \): Vertical shift \( k = 3 \), so horizontal asymptote \( y = 3 \).

Decay functions:

• \( z(x)=-5(2)^{x} \): No vertical shift ( \( k = 0 \)), horizontal asymptote \( y = 0 \).

• \( g(x)=-5(2)^{x}+3 \):