To analyze the end behavior of the function \( f(x) = \frac{8x + 1}{2x - 9} \), we can find the limits as \( x \) approaches \( -\infty \) and \( +\infty \).
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As \( x \to \infty \): \[ f(x) = \frac{8x + 1}{2x - 9} \] We can divide the numerator and the denominator by \( x \): \[ = \frac{8 + \frac{1}{x}}{2 - \frac{9}{x}} \] As \( x \to \infty \), \( \frac{1}{x} \to 0 \) and \( \frac{9}{x} \to 0 \): \[ \lim_{x \to \infty} f(x) = \frac{8 + 0}{2 - 0} = \frac{8}{2} = 4 \]
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As \( x \to -\infty \): Using the same approach: \[ f(x) = \frac{8x + 1}{2x - 9} = \frac{8 + \frac{1}{x}}{2 - \frac{9}{x}} \] Again, as \( x \to -\infty \), \( \frac{1}{x} \to 0 \) and \( \frac{9}{x} \to 0 \): \[ \lim_{x \to -\infty} f(x) = \frac{8 + 0}{2 - 0} = \frac{8}{2} = 4 \]
Thus, the end behavior of the function is:
- As \( x \to -\infty, f(x) \to 4 \)
- As \( x \to \infty, f(x) \to 4 \)
Therefore, the correct answer is: As \( x \to -\infty, f(x) \to 4; \) as ( x \to \infty, f(x) \to 4.
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