Which graph shows the transformation of the function f left parenthesis x right parenthesis equals e superscript x baseline where the function is translated three units to the right, vertically compressed by a factor of one fourth, and then translated six units down?

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Responses

A graph of an exponential function begins in quadrant 3, has a horizontal asymptote at y = negative 6, and passes through the points left-parenthesis negative 3 comma negative 2 right-parenthesis and left-parenthesis negative 2 comma 4.87 right-parenthesis.
Image with alt text: A graph of an exponential function begins in quadrant 3, has a horizontal asymptote at y = negative 6, and passes through the points left-parenthesis negative 3 comma negative 2 right-parenthesis and left-parenthesis negative 2 comma 4.87 right-parenthesis.

A graph of an exponential function begins in quadrant 3, has a horizontal asymptote at y = negative 6, and passes through the points left-parenthesis negative 3 comma negative 5.75 right-parenthesis and left-parenthesis negative 2 comma negative 5.32 right-parenthesis.
Image with alt text: A graph of an exponential function begins in quadrant 3, has a horizontal asymptote at y = negative 6, and passes through the points left-parenthesis negative 3 comma negative 5.75 right-parenthesis and left-parenthesis negative 2 comma negative 5.32 right-parenthesis.

A graph of an exponential function begins in quadrant 3, has a horizontal asymptote at y = negative 6, and passes through the points left-parenthesis 4 comma negative 5.32 right-parenthesis and left-parenthesis 5 comma negative 4.15 right-parenthesis.
Image with alt text: A graph of an exponential function begins in quadrant 3, has a horizontal asymptote at y = negative 6, and passes through the points left-parenthesis 4 comma negative 5.32 right-parenthesis and left-parenthesis 5 comma negative 4.15 right-parenthesis.

A graph of an exponential function begins in quadrant 3, has a horizontal asymptote at y = negative 6, and passes through the points left-parenthesis 4 comma 4.87 right-parenthesis and left-parenthesis 5 comma 23.56 right-parenthesis.
Image with alt text: A graph of an exponential function begins in quadrant 3, has a horizontal asymptote at y = negative 6, and passes through the points left-parenthesis 4 comma 4.87 right-parenthesis and left-parenthesis 5 comma 23.56 right-parenthesis.
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1 answer

To find the transformed function and determine which graph matches it, we start with the original function \( f(x) = e^x \).

Step 1: Translate 3 units to the right

Translating the function 3 units to the right involves replacing \( x \) with \( x - 3 \): \[ f(x) = e^{x - 3} \]

Step 2: Vertically compress by a factor of \( \frac{1}{4} \)

Next, we compress the function vertically by a factor of \( \frac{1}{4} \): \[ f(x) = \frac{1}{4} e^{x - 3} \]

Step 3: Translate 6 units down

Finally, we translate the function downwards by 6 units: \[ f(x) = \frac{1}{4} e^{x - 3} - 6 \]

Analyze the transformed function:

  1. Horizontal Asymptote: The horizontal asymptote of the function occurs where \( y \) approaches a constant value as \( x \) approaches infinity. For the current function, this asymptote is: \[ y = -6 \]

  2. Behavior of the function: The exponential function \( e^{x - 3} \) translates to the right, and the vertical compression makes it grow slower. The \( -6 \) shifts everything down.

Now, check the graphs:

  • All the graphs mentioned have a horizontal asymptote at \( y = -6 \).
  • Let's look for the behavior of the function. A vertically compressed exponential function shifted to the right will not have positive y-values near \( x = 3 \).

To check the points through which the transformed function passes:

  • If we evaluate at \( x = 3 \): \[ f(3) = \frac{1}{4} e^{3 - 3} - 6 = \frac{1}{4} - 6 = -\frac{23}{4} \approx -5.75 \]

  • If we evaluate at \( x = 4 \): \[ f(4) = \frac{1}{4} e^{4 - 3} - 6 = \frac{1}{4} e^1 - 6 \approx 0.67 - 6 = -5.33 \]

From the numerical approximations:

  • The y-values were approximately \( -5.75 \) for \( x = 3 \) and \( -5.33 \) for \( x = 4 \).

Conclusion:

The second graph matches the behavior of our transformed function, as it passes through points near \( (-3, -5.75) \) and \( (-2, -5.32) \), and has the correct horizontal asymptote.

Thus, the correct graph is:

  • A graph of an exponential function that begins in quadrant 3, has a horizontal asymptote at \( y = -6 \), and passes through the points \( \left(-3, -5.75\right) \) and \( \left(-2, -5.32\right) \).
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