Let \( f(x)=(x-6)^{6}(x-2)^{8}\left(2 x^{2}-x-5\right)^{3} \) \( \begin{array}{l}\text { As } x \rightarrow \infty, f(x) \rightarrow \\ -\infty \\ \text { Cannot Be Determined } \\ \text { As } x \rightarrow-\infty, f(x) \rightarrow \\ -\infty \\ \text { Cannot Be Determined }\end{array} \)
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As \( x \) approaches infinity, the leading term in \( f(x) \) is dominant, which is from \( (x-2)^8 \) since it has the highest exponent among \( (x-6)^6 \), \( (x-2)^8 \), and the quadratic term. Because all the terms with even powers will contribute positively, \( f(x) \rightarrow +\infty \) as \( x \rightarrow \infty \). When considering \( x \rightarrow -\infty \), the leading behavior will be influenced by the negative sign of the polynomial terms. As such, the term \( (x-6)^6 \) remains positive, \( (x-2)^8 \) stays positive, but the quadratic part \( (2x^2 - x - 5)^3 \) will behave according to the leading coefficient, \( 2x^2 \), which dominates and results in \( f(x) \rightarrow -\infty \). Thus, \( f(x) \) approaches negative infinity for large negative values of \( x \).