Which graph shows \( f(x)=3 \cdot 2^{x}-2 \) and its translation \( g(x)=3 \cdot 2^{x}+3 ? \)

1. We start with the function \[ f(x)=3\cdot 2^x-2. \] Its horizontal asymptote is determined by the constant term after the exponential part vanishes. As \(x\to -\infty\), \[ 3\cdot 2^x\to 0, \] so the horizontal asymptote is \[ y=-2. \] 2. Next, consider the translation given by \[ g(x)=3\cdot 2^x+3. \] Notice that compared to \(f(x)\), the only change is that the constant \(-2\) is replaced by \(+3\). This is equivalent to shifting \(f(x)\) vertically upward by 5 units (since \(3-(-2)=5\)). Therefore, \(g(x)\) is the graph of \(f(x)\) translated upward by 5. 3. The horizontal asymptote of \(g(x)\) is determined similarly. As \(x\to -\infty\), \[ 3\cdot 2^x\to 0, \] so the horizontal asymptote of \(g(x)\) is \[ y=3. \] 4. To identify the graphs: - The graph of \(f(x)=3\cdot 2^x-2\) is an exponential curve that approaches the line \(y=-2\) as \(x\to -\infty\) and increases rapidly as \(x\to\infty\). - The graph of \(g(x)=3\cdot 2^x+3\) is an exponential curve with the same shape as \(f(x)\) but shifted upward by 5 units. Its horizontal asymptote is \(y=3\). 5. Additionally, if we check a couple of key points: - For \(f(x)\): \( f(0)=3\cdot 2^0-2=3-2=1, \) \( f(1)=3\cdot 2^1-2=6-2=4. \) - For \(g(x)\): \( g(0)=3\cdot 2^0+3=3+3=6, \) \( g(1)=3\cdot 2^1+3=6+3=9. \) 6. Therefore, the correct graph will display: - Two exponential curves with the same shape. - One curve with a horizontal asymptote at \(y=-2\) going through \((0,1)\) and \((1,4)\) for \(f(x)\). - The other curve, which is just \(f(x)\) shifted upward by 5, having a horizontal asymptote at \(y=3\) and passing through \((0,6)\) and \((1,9)\) for \(g(x)\). The graph that shows these features—exponential curves with the described key points and asymptotes—is the one that correctly represents the functions \(f(x)=3\cdot 2^x-2\) and its vertical translation \(g(x)=3\cdot 2^x+3\).