Let $$ f(x)=(x-6)^{6}(x-2)^{8}\left(2 x^{2}-x-5 ight)^{3} $$ $$ \begin{array}{l} ext { As } x ightarrow \infty, f(x) ightarrow \ -\infty \ ext { Cannot Be Determined } \ ext { As } x ightarrow-\infty, f(x) ightarrow \ -\infty \ ext { Cannot Be Determined }\end{array} $$

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Let $$ f(x)=(x-6)^{6}(x-2)^{8}\left(2 x^{2}-x-5ight)^{3} $$ $$ \begin{array}{l}	ext { As } x ightarrow \infty, f(x) ightarrow \ -\infty \ 	ext { Cannot Be Determined } \ 	ext { As } x ightarrow-\infty, f(x) ightarrow \ -\infty \ 	ext { Cannot Be Determined }\end{array} $$

UpStudy AI · Accepted Answer

To analyze the behavior of the function $$ f(x) = (x-6)^{6}(x-2)^{8}(2x^{2}-x-5)^{3} $$ as $$ x $$ approaches $$ \infty $$ and $$ -\infty $$, we will consider the degree and leading coefficients of each factor in the function.

1. **As $$ x \rightarrow \infty $$**:
   - The term $$ (x-6)^{6} $$ behaves like $$ x^{6} $$ since the leading term dominates as $$ x $$ becomes very large.
   - The term $$ (x-2)^{8} $$ behaves like $$ x^{8} $$.
   - The term $$ (2x^{2}-x-5)^{3} $$ behaves like $$ (2x^{2})^{3} = 8x^{6} $$.

Now, we can combine these leading behaviors:
   $$
   f(x) \sim x^{6} \cdot x^{8} \cdot 8x^{6} = 8x^{20}
   $$
   As $$ x \rightarrow \infty $$, $$ f(x) \rightarrow \infty $$.

2. **As $$ x \rightarrow -\infty $$**:
   - The term $$ (x-6)^{6} $$ behaves like $$ (-x)^{6} = x^{6} $$ (since the exponent is even).
   - The term $$ (x-2)^{8} $$ behaves like $$ (-x)^{8} = x^{8} $$ (also even).
   - The term $$ (2x^{2}-x-5)^{3} $$ behaves like $$ (2x^{2})^{3} = 8x^{6} $$ (the leading term dominates).

Combining these leading behaviors:
   $$
   f(x) \sim x^{6} \cdot x^{8} \cdot 8x^{6} = 8x^{20}
   $$
   As $$ x \rightarrow -\infty $$, $$ f(x) \rightarrow \infty $$.

### Conclusion:
- As $$ x \rightarrow \infty $$, $$ f(x) \rightarrow \infty $$.
- As $$ x \rightarrow -\infty $$, $$ f(x) \rightarrow \infty $$.

Thus, the answers to the questions are:
- As $$ x \rightarrow \infty, f(x) \rightarrow \infty $$.
- As $$ x \rightarrow -\infty, f(x) \rightarrow \infty $$.
As $$ x $$ approaches both $$ \infty $$ and $$ -\infty $$, $$ f(x) $$ approaches $$ \infty $$.

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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