What are the asymptotic discontinuities of the given rational function below? $$ f(x)=\frac{-4}{x^{2}+11 x+10} $$ $$ x=-1 $$ only $$ x=-10 $$ only $$ x=-1 $$ and $$ x=-10 $$ $$ x=-5 $$

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To determine the vertical asymptotes (asymptotic discontinuities) of the rational function:

$$ f(x) = \frac{-4}{x^2 + 11x + 10} $$

we need to find the values of $$ x $$ that make the denominator zero, provided that the numerator is not zero at those points.

1. **Find the roots of the denominator:**

$$ x^2 + 11x + 10 = 0 $$

Using the quadratic formula:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
   
   Here, $$ a = 1 $$, $$ b = 11 $$, and $$ c = 10 $$.

$$ x = \frac{-11 \pm \sqrt{121 - 40}}{2} $$
   $$ x = \frac{-11 \pm \sqrt{81}}{2} $$
   $$ x = \frac{-11 \pm 9}{2} $$

This gives two solutions:
   
   $$ x = \frac{-11 + 9}{2} = -1 $$
   $$ x = \frac{-11 - 9}{2} = -10 $$

2. **Check the numerator at these points:**

The numerator is $$-4$$, which is never zero. Therefore, both $$ x = -1 $$ and $$ x = -10 $$ are vertical asymptotes of the function.

**Conclusion:**

The function $$ f(x) = \frac{-4}{x^2 + 11x + 10} $$ has vertical asymptotes at $$ x = -1 $$ and $$ x = -10 $$.

**Answer:**  
$$ x=-1 $$ and $$ x=-10 $$
The function has vertical asymptotes at $$ x = -1 $$ and $$ x = -10 $$.

What are the asymptotic discontinuities of the given rational function below? \( f(x)=\frac{-4}{x^{2}+11 x+10} \) \( x=-1 \) only \( x=-10 \) only \( x=-1 \) and \( x=-10 \) \( x=-5 \)

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