To find the transformed function from \( f(x) = e^x \) after applying the specified transformations, we'll follow these steps:
- Translate 3 units to the right: This transforms the function to \( f(x - 3) = e^{(x-3)} \).
- Vertically compress by a factor of \(\frac{1}{4}\): This changes the function to \( \frac{1}{4} e^{(x-3)} \).
- Translate 6 units downward: This results in \( \frac{1}{4} e^{(x-3)} - 6 \).
The final transformation is: \[ g(x) = \frac{1}{4} e^{(x-3)} - 6. \]
Now, we can analyze the properties of this transformed function:
-
Horizontal Asymptote: As \( x \) approaches infinity, \( \frac{1}{4} e^{(x-3)} \) goes to infinity. However, as \( x \) approaches negative infinity, the exponential term approaches \( 0 \). Therefore: \[ g(x) \to 0 - 6 = -6. \] Thus, the horizontal asymptote is \( y = -6 \).
-
Behavior of the function: As \( x \) increases, the function will start below -6 and approach -6 as \( x \) moves to negative infinity, but will eventually rise towards infinity.
Now, let's check the options provided:
-
The first graph has points \((-3, -2)\) and \((-2, 4.87)\) and does not seem compatible with our derived asymptote.
-
The second graph has points \((-3, -5.75)\) and \((-2, -5.32)\). This graph appears to align with the behavior of having a horizontal asymptote at \(y = -6\) and could represent the transformation as it is in quadrant 3.
-
The third graph has points \((4, -5.32)\) and \((5, -4.15)\), which indicates it is above the horizontal asymptote at \(y = -6\), thus not representing our function.
-
The fourth graph includes points \((4, 4.87)\) and \((5, 23.56)\), clearly suggesting the function has moved too high, thus also not aligning with our derived function.
Conclusion: The second graph, which has a horizontal asymptote at \(y = -6\) and passes through points like \((-3, -5.75)\) and \((-2, -5.32)\), is the best match for the described transformation of \(f(x)\).
1 answer