Graphs of Exponential Functions

Problem: Let function \( f \) be given by \( f(x) = 2^{x - 2} \).

1. Find the domain and range of \( f \).
2. Find the horizontal asymptote of the graph.
3. Find the \( x \)- and \( y \)-intercepts of the graph of \( f \) if there are any.
4. Sketch the graph of \( f \).

Solution:

1. The domain of \( f \) is the set of all real numbers \( (-\infty, +\infty) \).
   To find the range of \( f \), we start with: \[ 2^x \gt 0 \] Multiply both sides by \( 2^{-2} \) (a positive number): \[ 2^x \cdot 2^{-2} \gt 0 \implies 2^{x-2} \gt 0 \] This inequality indicates that \( f(x) \gt 0 \). Hence, the range is given by the interval: \[ (0, +\infty) \]
2. As \( x \) decreases without bound (\( x \to -\infty \)), \( f(x) = 2^{x-2} \) approaches \( 0 \). Hence, the graph of \( f \) has a horizontal asymptote at \( y = 0 \).
3. To find the \( x \)-intercept, we solve \( f(x) = 0 \): \[ 2^{x-2} = 0 \] Since \( 2^{x-2} \gt 0 \) for all real numbers \( x \), this equation has no solution. The graph of \( f \) does not have an \( x \)-intercept.
   The \( y \)-intercept is given by: \[ (0, f(0)) = (0, 2^{0-2}) = \left(0, \frac{1}{4}\right) \]
4. To sketch the graph accurately, we plot additional coordinates: \[ (4, f(4)) = (4, 2^{4-2}) = (4, 2^2) = (4, 4) \] \[ (-1, f(-1)) = (-1, 2^{-1-2}) = (-1, 2^{-3}) = \left(-1, \frac{1}{8}\right) \]

   

Problem: Function \( f \) is given by \( f(x) = 2^{x + 2} \).

1. Find the domain and range of \( f \).
2. Find the horizontal asymptote of the graph.
3. Find the \( x \)- and \( y \)-intercepts of the graph of \( f \) if there are any.
4. Sketch the graph of \( f \).

Problem: Let \( f \) be a function given by \( f(x) = 3^{x+1} - 2 \).

1. Find the domain and range of \( f \).
2. Find the horizontal asymptote of the graph.
3. Find the \( x \)- and \( y \)-intercepts of the graph of \( f \) if there are any.
4. Sketch the graph of \( f \).

Solution:

1. The domain of \( f \) is all real numbers \( (-\infty, +\infty) \).
   To find the range, start with: \[ 3^x \gt 0 \] Multiply both sides by \( 3 \) (positive): \[ 3 \cdot 3^x \gt 3 \cdot 0 \implies 3^{x+1} \gt 0 \] Subtract \( 2 \) from both sides: \[ 3^{x+1} - 2 \gt -2 \] Thus, \( f(x) \gt -2 \). The range is: \[ (-2, +\infty) \]
2. As \( x \) decreases without bound (\( x \to -\infty \)), \( 3^{x+1} \) approaches \( 0 \), so \( f(x) = 3^{x+1} - 2 \) approaches \( -2 \). The horizontal asymptote is \( y = -2 \).
3. To find the \( x \)-intercept, solve \( f(x) = 0 \): \[ 3^{x+1} - 2 = 0 \implies 3^{x+1} = 2 \] Rewrite in logarithmic form and solve for \( x \): \[ x + 1 = \log_3(2) \implies x = \log_3(2) - 1 \] The \( y \)-intercept is given by: \[ (0, f(0)) = (0, 3^{0+1} - 2) = (0, 3 - 2) = (0, 1) \]
4. Calculating extra reference points for sketching: \[ (-2, f(-2)) = (-2, 3^{-2+1} - 2) = \left(-2, \frac{1}{3} - 2\right) \approx (-2, -1.67) \] \[ (-4, f(-4)) = (-4, 3^{-4+1} - 2) = \left(-4, \frac{1}{27} - 2\right) \approx (-4, -1.96) \]

   

Problem: Function \( f \) given by \( f(x) = 2^{x - 2} + 1 \).

1. Find the domain and range of \( f \).
2. Find the horizontal asymptote of the graph of \( f \).
3. Find the \( x \)- and \( y \)-intercepts of the graph of \( f \) if there are any.
4. Sketch the graph of \( f \).