Question

v(x) = 3(5)^x - 3
n(x) = 5(3)^x + 3
z(x) = 5(3)^x - 3
g(x) = 3(5)^x + 3

Explanation:

Step1: Analyze \( g(x) = 3(5)^x + 3 \)

The parent function is \( y = 5^x \), vertically stretched by 3 (\( 3(5)^x \)) and shifted up 3 units (\( +3 \)). At \( x = 0 \), \( g(0)=3(1)+3 = 6 \). The y - intercept is 6, and the horizontal asymptote is \( y = 3 \) (since exponential functions of the form \( a\cdot b^x + k \) have horizontal asymptote \( y = k \)). Looking at the graphs, the top - left graph has a y - intercept around 9? Wait, no, let's recalculate. Wait, \( g(x)=3(5)^x + 3 \), when \( x = 0 \), \( 3\times1+3 = 6 \). Wait, maybe I made a mistake. Wait, the top - left graph: let's check the y - intercept. The top - left graph's curve at \( x = 0 \) is around \( y = 9 \)? No, wait the first function \( v(x)=3(5)^x - 3 \), at \( x = 0 \), \( 3 - 3=0 \). \( n(x)=5(3)^{x + 3} \), at \( x=-3 \), \( 5(3)^0 = 5 \), and it's a vertical shift? Wait, no, \( n(x)=5(3)^{x+3}=5\times3^3\times3^x=135\times3^x \), so it's a vertical stretch. \( z(x)=5(3)^{x - 3}=\frac{5}{27}\times3^x \), vertical compression. \( g(x)=3(5)^x+3 \), at \( x = 0 \), \( 3 + 3=6 \), horizontal asymptote \( y = 3 \).

Wait, let's list the y - intercepts:

- \( v(x)=3(5)^x - 3 \): \( x = 0 \), \( 3(1)-3 = 0 \)
- \( n(x)=5(3)^{x + 3} \): \( x = 0 \), \( 5(3)^3=5\times27 = 135 \) (very large y - intercept)
- \( z(x)=5(3)^{x - 3} \): \( x = 0 \), \( 5(3)^{-3}=\frac{5}{27}\approx0.185 \)
- \( g(x)=3(5)^x + 3 \): \( x = 0 \), \( 3(1)+3 = 6 \)

Now, looking at the graphs:

Top - left graph: y - intercept around 9? No, wait the first graph (top - left) has a curve that starts near \( y = 3 \) (horizontal asymptote) and goes up, with y - intercept around 9? Wait, maybe I miscalculated \( g(x) \). Wait \( 3(5)^x \) when \( x = 0 \) is 3, plus 3 is 6. Wait the top - left graph: let's check the horizontal asymptote. The top - left graph's horizontal asymptote is around \( y = 3 \)? No, the top - left graph's curve is above \( y = 3 \), and at \( x = 0 \), it's around \( y = 9 \)? Wait, no, maybe the first function \( g(x) \) is the top - left? Wait, no, let's take \( g(x)=3(5)^x + 3 \). The general form of an exponential function \( y = a\cdot b^x + k \) has horizontal asymptote \( y = k \). For \( g(x) \), \( k = 3 \), so the horizontal asymptote is \( y = 3 \). The top - left graph has a horizontal asymptote around \( y = 3 \) (the curve approaches a line around \( y = 3 \) as \( x\to-\infty \)) and at \( x = 0 \), \( y\approx9 \)? Wait, no, \( 3(5)^x+3 \), when \( x = 1 \), \( 3\times5 + 3=18 \), \( x = 0 \), \( 6 \), \( x=-1 \), \( 3\times\frac{1}{5}+3=\frac{3}{5}+3 = 3.6 \). So the graph should have a horizontal asymptote at \( y = 3 \), pass through (0,6), and increase rapidly. The top - left graph: let's see the grid. Each square is 1 unit. The top - left graph's curve at \( x = 0 \) is at \( y = 9 \)? No, maybe I mixed up the functions.

Wait, let's take \( v(x)=3(5)^x - 3 \). Horizontal asymptote \( y=-3 \), at \( x = 0 \), \( y = 0 \). The bottom - left graph: at \( x = 0 \), \( y = 0 \), and horizontal asymptote \( y=-3 \)? No, the bottom - left graph's horizontal asymptote is \( y = 0 \)? Wait, no. Wait the bottom - right graph: let's check. The bottom - right graph has a horizontal asymptote around \( y=-3 \)? Wait, \( v(x)=3(5)^x - 3 \), horizontal asymptote \( y=-3 \), and at \( x = 0 \), \( y = 0 \). So the bottom - right graph: when \( x\to-\infty \), the curve approaches \( y=-3 \), and at \( x = 0 \), \( y = 0 \). That matches \( v(x)=3(5)^x - 3 \).

\( g(x)=3(5)^x + 3 \): horizontal asymptote \( y = 3 \), at \( x = 0 \), \( y = 6 \). The top - left graph: when \( x\to-\infty \), approaches \( y = 3 \), and at \( x = 0 \), \( y\approx9 \)? No, wait \( 3(5)^x+3 \), at \( x = 1 \), \( 15 + 3=18 \), so it's increasing rapidly. The top - left graph: let's see, the curve is increasing, horizontal asymptote \( y = 3 \) (as \( x\to-\infty \)), and at \( x = 0 \), \( y = 6 \)? Wait, maybe the top - left graph is \( g(x) \).

\( n(x)=5(3)^{x + 3} \): this is a very steep curve because the base is 3 and it's multiplied by 5 and shifted left 3 units. So at \( x = 0 \), \( y = 135 \), which would be a graph that shoots up very quickly, like the top - right graph.

\( z(x)=5(3)^{x - 3} \): this is a curve that is compressed vertically, at \( x = 0 \), \( y=\frac{5}{27}\approx0.185 \), and increases, with horizontal asymptote \( y = 0 \). The bottom - left graph: at \( x = 0 \), \( y = 0 \)? No, the bottom - left graph at \( x = 0 \), \( y = 0 \), and horizontal asymptote \( y = 0 \)? Wait, no, \( z(x)=5(3)^{x - 3} \) has horizontal asymptote \( y = 0 \), and at \( x = 3 \), \( y = 5 \). So the bottom - left graph: when \( x = 3 \), \( y = 5 \), and it's increasing. Wait, maybe I need to re - approach.

Let's match each function to the graph:

1. \( v(x)=3(5)^x - 3 \):
- Horizontal asymptote: \( y=-3 \) (since \( \lim_{x\to-\infty}3(5)^x-3=-3 \))
- y - intercept: \( x = 0 \), \( 3(5)^0-3 = 0 \)
- The graph that has horizontal asymptote \( y=-3 \) and y - intercept 0 is the bottom - right graph.

2. \( g(x)=3(5)^x + 3 \):
- Horizontal asymptote: \( y = 3 \) (since \( \lim_{x\to-\infty}3(5)^x+3 = 3 \))
- y - intercept: \( x = 0 \), \( 3(5)^0+3 = 6 \)
- The graph that has horizontal asymptote \( y = 3 \) and y - intercept around 6 (top - left graph, as it approaches \( y = 3 \) as \( x\to-\infty \) and has a y - intercept around 9? Wait, maybe my calculation of the y - intercept is wrong. Wait \( 3(5)^x+3 \), when \( x = 0 \), it's \( 3 + 3=6 \), but the top - left graph's y - intercept is at \( y = 9 \)? Wait, maybe the first function \( v(x)=3(5)^x - 3 \) has y - intercept 0, and the bottom - right graph has y - intercept 0 and horizontal asymptote \( y=-3 \), which matches.

3. \( n(x)=5(3)^{x + 3} \):
- This is \( 5\times3^{x+3}=5\times3^3\times3^x = 135\times3^x \), so it's a vertically stretched exponential function with a large coefficient. At \( x = 0 \), \( y = 135 \), so it will be a very steep curve, like the top - right graph (which goes up very quickly).

4. \( z(x)=5(3)^{x - 3} \):
- This is \( 5\times3^{x - 3}=\frac{5}{27}\times3^x \), a vertically compressed exponential function. At \( x = 0 \), \( y=\frac{5}{27}\approx0.185 \), and as \( x \) increases, it grows. The bottom - left graph: at \( x = 0 \), \( y = 0 \) (approximate) and has a horizontal asymptote \( y = 0 \), which matches \( z(x)=5(3)^{x - 3} \) (since \( \lim_{x\to-\infty}\frac{5}{27}\times3^x = 0 \)).

Step2: Match each function to the graph

- \( v(x)=3(5)^x - 3 \): Bottom - right graph (horizontal asymptote \( y=-3 \), y - intercept 0)
- \( g(x)=3(5)^x + 3 \): Top - left graph (horizontal asymptote \( y = 3 \), y - intercept around 6 - 9? Wait, maybe my earlier y - intercept calculation was wrong. Wait \( 3(5)^x+3 \), when \( x = 1 \), \( 15 + 3=18 \), so the top - left graph, which has a steep increase and horizontal asymptote \( y = 3 \), is \( g(x) \).
- \( n(x)=5(3)^{x + 3} \): Top - right graph (very steep, large y - intercept)
- \( z(x)=5(3)^{x - 3} \): Bottom - left graph (vertically compressed, small y - intercept, horizontal asymptote \( y = 0 \))

Answer:

To match the functions to the graphs:

• \( v(x)=3(5)^x - 3 \): Bottom - right graph
• \( g(x)=3(5)^x + 3 \): Top - left graph
• \( n(x)=5(3)^{x + 3} \): Top - right graph
• \( z(x)=5(3)^{x - 3} \): Bottom - left graph

(If the question was to match a specific function, for example, if we take \( v(x)=3(5)^x - 3 \), the answer is the bottom - right graph; for \( g(x)=3(5)^x + 3 \), top - left; for \( n(x)=5(3)^{x + 3} \), top - right; for \( z(x)=5(3)^{x - 3} \), bottom - left)

Assuming we are matching \( v(x)=3(5)^x - 3 \), the answer is the bottom - right graph.