Which graph shows \( f(x)=3 \cdot 2^{x-7}+3 \) and its translation \( g(x)=f(x+7) \)

1. We start with \[ f(x)=3\cdot 2^{\,x-7}+3. \] Note that compared to the “base‐exponential” function, this function has been shifted horizontally and vertically. In particular, the exponent \(x-7\) indicates a rightward shift by 7 units, and the “\(+3\)” at the end translates the graph upward by 3 units. Its horizontal asymptote is at \(y=3\) since \[ \lim_{x\to-\infty} 3\cdot 2^{\,x-7}+3 = 3. \] 2. Next, we consider the translated function \[ g(x)=f(x+7). \] Substituting \(x+7\) into \(f\) yields \[ g(x)=3\cdot 2^{\,(x+7)-7}+3 = 3\cdot 2^{\,x}+3. \] Hence, \(g(x)\) is simply the function \[ g(x)=3\cdot 2^x+3. \] 3. Comparing the two functions: - The graph of \(f(x)=3\cdot 2^{x-7}+3\) has a horizontal asymptote \(y=3\) and, due to the shift \(x-7\), passes through the point where the exponent equals zero. Setting \(x-7=0\) gives \(x=7\), and at this point \[ f(7)=3\cdot 2^0+3=3\cdot 1+3=6. \] So the point \((7,6)\) lies on the graph of \(f\). - The graph of \(g(x)=3\cdot 2^x+3\) has the same vertical shift (asymptote at \(y=3\)), but its “critical” point now occurs when \(x=0\) since the exponent is \(x\). Indeed, \[ g(0)=3\cdot 2^0+3=3+3=6. \] Thus, the point \((0,6)\) lies on \(g\). 4. The translation from \(f\) to \(g\) is exactly a horizontal shift of 7 units to the left. In other words, every point on the graph of \(f\) is moved 7 units left to obtain the graph of \(g\). 5. Therefore, the correct graph shows: - The curve for \(f(x)=3\cdot 2^{x-7}+3\) with horizontal asymptote \(y=3\) and passing through \((7,6)\), - And the translated curve for \(g(x)=f(x+7)=3\cdot 2^x+3\) with the same asymptote and passing through \((0,6)\). Any graph that displays an exponential curve with these properties – the original curve shifted 7 units right with \((7,6)\) on it and its translation shifted 7 units left so that it passes through \((0,6)\) – is the correct one.