To find the asymptotes of the hyperbola given by the equation:
\[ \frac{x^2}{16} - \frac{(y-4)^2}{7} = 1, \]
we can start by identifying the standard form of the hyperbola. The general form of a horizontal hyperbola centered at \((h, k)\) is given by:
\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1. \]
In our case, we have:
- \( a^2 = 16 \) \( \Rightarrow a = 4 \)
- \( b^2 = 7 \) \( \Rightarrow b = \sqrt{7} \)
- The center \((h, k) = (0, 4)\).
The equations of the asymptotes for a hyperbola centered at \((h, k)\) are given by:
\[ y - k = \pm \frac{b}{a}(x - h). \]
For our hyperbola, substituting the center \((0, 4)\), we have:
\[ y - 4 = \pm \frac{\sqrt{7}}{4}(x - 0) \]
This gives us the equations of the asymptotes:
- \( y - 4 = \frac{\sqrt{7}}{4}x \) \(\Rightarrow y = \frac{\sqrt{7}}{4}x + 4\)
- \( y - 4 = -\frac{\sqrt{7}}{4}x \) \(\Rightarrow y = -\frac{\sqrt{7}}{4}x + 4\)
These asymptotes represent two lines that approach the branches of the hyperbola as \(x\) moves towards positive and negative infinity.
Now, let's analyze the given options:
- \( y = 4 \) is a horizontal line through the center but is not an asymptote.
- \( y = -4 \) is also not related to the hyperbola's asymptotes.
- \( x = 0 \) is a vertical line through the center, but it’s not an asymptote either.
- \( y = 0 \) is also not relevant as an asymptote.
None of the provided options correspond to the asymptotes derived from our hyperbola equation.
Thus, the asymptotes for this hyperbola are:
- \( y = \frac{\sqrt{7}}{4}x + 4 \)
- \( y = -\frac{\sqrt{7}}{4}x + 4 \)
None of the provided options correctly represent the asymptotes.