The problem asks us to determine what types of asymptotes the function $f(x) = \frac{3x^3 + 2x^2 + x + 1}{x^2 + x - 2}$ has.

1. Problem Description

The problem asks us to determine what types of asymptotes the function $f(x) = \frac{3x^3 + 2x^2 + x + 1}{x^2 + x - 2}$ has.

2. Solution Steps

First, we determine the vertical asymptotes. Vertical asymptotes occur where the denominator of a rational function is zero and the numerator is nonzero. We set the denominator equal to zero and solve for $x$:

$x^2 + x - 2 = 0$

$(x + 2)(x - 1) = 0$

$x = -2$ or $x = 1$

We check if the numerator is zero at these values.

For $x = -2$, $3(-2)^3 + 2(-2)^2 + (-2) + 1 = 3(-8) + 2(4) - 2 + 1 = -24 + 8 - 2 + 1 = -17 \neq 0$.

For $x = 1$, $3(1)^3 + 2(1)^2 + (1) + 1 = 3 + 2 + 1 + 1 = 7 \neq 0$.

Therefore, there are vertical asymptotes at $x = -2$ and $x = 1$.

Next, we consider horizontal and oblique asymptotes. A horizontal asymptote exists if the degree of the numerator is less than or equal to the degree of the denominator. An oblique asymptote exists if the degree of the numerator is exactly one more than the degree of the denominator. In this case, the degree of the numerator is 3, and the degree of the denominator is

2. Since the degree of the numerator is one more than the degree of the denominator, there is an oblique asymptote, and there is no horizontal asymptote.

To find the oblique asymptote, we perform polynomial long division:

```

3x - 1

x^2+x-2 | 3x^3+2x^2+x+1

-(3x^3+3x^2-6x)

----------------

-x^2+7x+1

-(-x^2-x+2)

-------------

8x-1

```

Thus, $f(x) = 3x - 1 + \frac{8x - 1}{x^2 + x - 2}$. As $x$ approaches infinity, $\frac{8x - 1}{x^2 + x - 2}$ approaches

0. Therefore, the oblique asymptote is $y = 3x - 1$.

Since there are vertical asymptotes at $x = -2$ and $x = 1$, and an oblique asymptote at $y = 3x - 1$, $f(x)$ has oblique and vertical asymptotes.

3. Final Answer

Oblique and vertical asymptotes