Write an equation of the slant or nonlinear asymptote. $$ r(x)=\frac{-x^{3}-13 x^{2}-35 x+3}{x+9} $$ The (Choose one) $$ \nabla $$ asymptote is $$ y=\square $$

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To find the slant or nonlinear asymptote of the rational function

$$
r(x) = \frac{-x^{3}-13 x^{2}-35 x+3}{x+9},
$$

we need to perform polynomial long division since the degree of the numerator is greater than the degree of the denominator.

### Step 1: Polynomial Long Division

1. Divide the leading term of the numerator $$-x^3$$ by the leading term of the denominator $$x$$ to get $$-x^2$$.
2. Multiply the entire denominator $$x + 9$$ by $$-x^2$$ and subtract from the numerator.
3. Repeat the process with the new polynomial obtained after subtraction.

Let's perform these steps.

1. **First Division:**
   $$
   -x^3 \div x = -x^2
   $$
   Multiply:
   $$
   -x^2(x + 9) = -x^3 - 9x^2
   $$
   Subtract:
   $$
   (-x^3 - 13x^2 - 35x + 3) - (-x^3 - 9x^2) = -4x^2 - 35x + 3
   $$

2. **Second Division:**
   $$
   -4x^2 \div x = -4x
   $$
   Multiply:
   $$
   -4x(x + 9) = -4x^2 - 36x
   $$
   Subtract:
   $$
   (-4x^2 - 35x + 3) - (-4x^2 - 36x) = x + 3
   $$

3. **Third Division:**
   $$
   x \div x = 1
   $$
   Multiply:
   $$
   1(x + 9) = x + 9
   $$
   Subtract:
   $$
   (x + 3) - (x + 9) = -6
   $$

### Step 2: Result of Long Division

The result of the long division is:
$$
r(x) = -x^2 - 4x + 1 + \frac{-6}{x + 9}
$$

### Step 3: Identify the Asymptote

As $$x$$ approaches infinity, the term $$\frac{-6}{x + 9}$$ approaches 0. Therefore, the slant asymptote is given by the polynomial part of the division:

$$
y = -x^2 - 4x + 1
$$

### Conclusion

The slant asymptote is:

$$
y = -x^2 - 4x + 1
$$

Thus, the answer is:

The $$ \nabla $$ asymptote is $$ y = -x^2 - 4x + 1 $$.
The slant asymptote is $$ y = -x^2 - 4x + 1 $$.

Write an equation of the slant or nonlinear asymptote. \[ r(x)=\frac{-x^{3}-13 x^{2}-35 x+3}{x+9} \] The (Choose one) \( \nabla \) asymptote is \( y=\square \)

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