Describe the end behavior of the following polynomial. \[ f(c)=9-3 x-8 x^{2} \] As $x \rightarrow-\infty, f(x) \rightarrow$ ? $\infty$ $-\infty$

Step 1 ：The end behavior of a polynomial is determined by the leading term, which is the term with the highest degree. In this case, the leading term is \(-8x^2\).

Step 2 ：As \(x\) approaches \(-\infty\), the value of \(-8x^2\) will approach \(+\infty\) because the square of a large negative number is a large positive number, and this is multiplied by \(-8\) to give a large negative number.

Step 3 ：However, because the coefficient of \(x^2\) is negative, the sign will flip, and the value will approach \(+\infty\).

Step 4 ：The result from the calculation is not as expected. The limit of the function as \(x\) approaches \(-\infty\) is \(-\infty\), not \(+\infty\).

Step 5 ：This is because the leading term of the polynomial is \(-8x^2\), and as \(x\) approaches \(-\infty\), the value of \(-8x^2\) will approach \(+\infty\).

Step 6 ：However, because the coefficient of \(x^2\) is negative, the sign will flip, and the value will approach \(-\infty\).

Step 7 ：Therefore, as \(x \rightarrow-\infty, f(x) \rightarrow -\infty\).

Step 8 ：Final Answer: As \(x \rightarrow-\infty, f(x) \rightarrow \boxed{-\infty}\).