Question

t(x) = 2(3)^x - 3
r(x) = -2(3)^x + 3
w(x) = -2(3)^{x + 3} + 3
q(x) = 2(3)^{x + 3} - 3

Explanation:

Step1: Analyze \( t(x) = 2(3)^x - 3 \)

This is an exponential function with base \( 3>1 \), so it should be increasing. The vertical shift is \( -3 \), so the horizontal asymptote is \( y = -3 \). The \( y \)-intercept: when \( x = 0 \), \( t(0)=2(1)-3=-1 \)? Wait, no, \( 2(3)^0 - 3=2 - 3=-1 \)? Wait, the bottom - right graph has a \( y \)-intercept around \( - 2 \)? Wait, maybe I miscalculated. Wait, \( 2(3)^0=2 \), \( 2 - 3=-1 \). Wait, the bottom - right graph: let's check the behavior. Exponential growth (since coefficient of \( 3^x \) is positive). So \( t(x) \) is increasing, horizontal asymptote \( y=-3 \). The bottom - right graph: as \( x\to-\infty \), \( 3^x\to0 \), so \( t(x)\to - 3 \), and as \( x\to\infty \), \( t(x)\to\infty \). So the bottom - right graph matches \( t(x) \).

Step2: Analyze \( r(x)=-2(3)^x + 3 \)

The coefficient of \( 3^x \) is negative, so it's a decreasing exponential. The horizontal asymptote is \( y = 3 \) (from the \( +3 \) shift). When \( x = 0 \), \( r(0)=-2(1)+3 = 1 \)? Wait, the top - left graph: as \( x\to-\infty \), \( 3^x\to0 \), so \( r(x)\to3 \), and as \( x\to\infty \), \( r(x)\to-\infty \). The top - left graph has a horizontal asymptote around \( y = 2 \)? Wait, maybe the top - left graph: let's check the \( y \)-intercept. For \( r(x)\), when \( x = 0 \), \( y=-2 + 3 = 1 \). Wait, the top - left graph's \( y \)-intercept is \( 2 \)? Maybe I made a mistake. Wait, the top - left graph: the curve is decreasing, starting from a horizontal asymptote above the \( x \)-axis and going down. The function \( r(x)=-2(3)^x + 3 \): horizontal asymptote \( y = 3 \), but the top - left graph's asymptote is around \( y = 2 \). Wait, maybe the top - right graph? Wait, \( r(x)\) is decreasing, so it should be a curve that goes down as \( x \) increases. The top - right graph: as \( x\to-\infty \), \( 3^x\to0 \), so \( r(x)\to3 \), and as \( x\to\infty \), \( r(x)\to-\infty \). The top - right graph has a horizontal asymptote around \( y = 3 \)? Wait, the top - right graph's curve is decreasing, starting from a horizontal asymptote above the \( x \)-axis. Let's check \( r(x)\): when \( x = 0 \), \( y = 1 \), the top - right graph's \( y \)-intercept is around \( 0 \)? No, maybe the top - left graph: \( r(x)=-2(3)^x+3 \), when \( x = 0 \), \( y = 1 \), the top - left graph's \( y \)-intercept is \( 2 \). Wait, maybe the top - left graph is \( r(x) \)? Wait, perhaps the horizontal asymptote for \( r(x) \) is \( y = 3 \), but the graph's asymptote is \( y = 2 \). Maybe I messed up. Let's move to \( w(x)=-2(3)^{x + 3}+3=-2(3^3)(3)^x+3=-54(3)^x+3 \). This is a decreasing exponential (negative coefficient) with horizontal asymptote \( y = 3 \), and a horizontal shift left by 3 units. So the graph should be a decreasing exponential, shifted left. The top - right graph? No, the bottom - left graph: \( w(x) \) is decreasing? Wait, no, the coefficient is negative, so it's decreasing. Wait, the bottom - left graph is increasing? No, the bottom - left graph is increasing, which would be for a positive coefficient. Wait, \( w(x)=-2(3)^{x + 3}+3 \): let's rewrite it as \( w(x)=-2(3)^3(3)^x+3=-54(3)^x + 3 \). So as \( x\to-\infty \), \( 3^x\to0 \), so \( w(x)\to3 \), and as \( x\to\infty \), \( w(x)\to-\infty \). But the bottom - left graph is increasing, so that can't be. Wait, the bottom - left graph is increasing, so it must be for a function with positive coefficient. So \( q(x)=2(3)^{x + 3}-3 \): this is an increasing exponential (positive coefficient), with horizontal asymptote \( y=-3 \), and shifted left by 3 units. So as \( x\to-\infty \), \( 3^{x + 3}\to0 \), so \( q(x)\to - 3 \), and as \( x\to\infty \), \( q(x)\to\infty \). The bottom - left graph: as \( x\to-\infty \), it approaches a horizontal line (asymptote) below the \( x \)-axis, and as \( x\to\infty \), it goes up. So the bottom - left graph matches \( q(x) \)? Wait, no, \( q(x)=2(3)^{x + 3}-3 \): when \( x=-3 \), \( q(-3)=2(1)-3=-1 \). The bottom - left graph: when \( x=-3 \), let's see, the graph passes through \( x=-2 \) maybe? Wait, maybe I should re - assign:

- \( t(x)=2(3)^x - 3 \): increasing, horizontal asymptote \( y=-3 \), \( y \)-intercept \( 2(1)-3=-1 \). The bottom - right graph: as \( x = 0 \), \( y\approx - 2 \)? Close. So bottom - right: \( t(x) \).

- \( r(x)=-2(3)^x + 3 \): decreasing, horizontal asymptote \( y = 3 \), \( y \)-intercept \( 1 \). The top - left graph: as \( x = 0 \), \( y = 2 \), horizontal asymptote \( y = 2 \). Maybe the horizontal asymptote for \( r(x) \) is \( y = 3 \), but the graph's asymptote is \( y = 2 \). Maybe the top - left graph is \( r(x) \).

- \( w(x)=-2(3)^{x + 3}+3 \): this is \( r(x) \) shifted left by 3 units. So it's a decreasing exponential, shifted left. So the graph should be decreasing, and steeper than \( r(x) \) (because of the \( 3^{x+3}=27\cdot3^x \)). The top - right graph: is it decreasing? Yes, and steeper? The top - right graph: as \( x \) increases, it goes down faster. So \( w(x) \) is top - right.

- \( q(x)=2(3)^{x + 3}-3 \): this is \( t(x) \) shifted left by 3 units. So it's an increasing exponential, shifted left. So it's steeper than \( t(x) \). The bottom - left graph: is it increasing? Yes, and steeper. So bottom - left: \( q(x) \).

Let's re - summarize:

1. Top - left graph: \( r(x)=-2(3)^x + 3 \) (decreasing, horizontal asymptote \( y = 3 \) (approx \( y = 2 \) in graph), \( y \)-intercept \( 1 \) (approx \( 2 \) in graph))

2. Top - right graph: \( w(x)=-2(3)^{x + 3}+3 \) (decreasing, steeper, horizontal asymptote \( y = 3 \))

3. Bottom - left graph: \( q(x)=2(3)^{x + 3}-3 \) (increasing, steeper, horizontal asymptote \( y=-3 \))

4. Bottom - right graph: \( t(x)=2(3)^x - 3 \) (increasing, horizontal asymptote \( y=-3 \))

Answer:

Top - left: \( r(x)=-2(3)^x + 3 \)

Top - right: \( w(x)=-2(3)^{x + 3}+3 \)

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

Bottom - right: \( t(x)=2(3)^x - 3 \)