Which of the following transformations results in a vertical stretch of the exponential decay function $f\left(x\right)=e^{-x}$?

• A. $g\left(x\right)=e^{-x}+3$
• B. $g\left(x\right)=3e^{-x}$
• C. $g\left(x\right)=e^{-3x}$
• D. $g\left(x\right)=\frac{1}{3}{e}^{-x}$

Answer: (B)

Verified step by step guidance

1. Step 1: Understand the concept of a vertical stretch. A vertical stretch occurs when the function is multiplied by a constant greater than 1. This amplifies the output values of the function, making it 'taller' or 'stretched' vertically.
2. Step 2: Analyze the given function f(x) = e^{-x}. This is an exponential decay function, where the base of the exponential is e and the exponent is -x.
3. Step 3: Examine the transformations provided: g(x) = e^{-x} + 3, g(x) = 3e^{-x}, g(x) = e^{-3x}, and g(x) = (1/3)e^{-x}. Note that adding a constant (e.g., +3) shifts the graph vertically but does not stretch it. Similarly, multiplying the exponent (e.g., -3x) affects the rate of decay but does not stretch the graph vertically.
4. Step 4: Focus on g(x) = 3e^{-x}. Here, the function f(x) = e^{-x} is multiplied by 3, which amplifies all output values by a factor of 3. This is a vertical stretch because the constant multiplier is greater than 1.
5. Step 5: Verify that g(x) = (1/3)e^{-x} represents a vertical compression, not a stretch, because the multiplier (1/3) is less than 1. Therefore, the correct transformation for a vertical stretch is g(x) = 3e^{-x}.