F(X) = 2x^3-5x^2-19x+1 / x^2-9....I need the vertical asymptote, horizontal asymptote and the slant asymptote...please help!

Vertical asymptote: denominator = 0, so: x^2 - 9 = 0

Horizontal: lim x->(infinity) f(x)
Since the degree of nominator is higher than the denominator, then the horizontal asymptote does not exist.

Slant: use long division method to find the quotient. That quotient is the slant asymptote
In this case: divide 2x^3-5x^2-19x+1 with x^2-9

k(x)= x^4-2x +7/x^2-9

To find the vertical asymptote, horizontal asymptote, and slant asymptote of the function f(x) = (2x^3 - 5x^2 - 19x + 1) / (x^2 - 9), we can use some rules and techniques from calculus.

1. Vertical Asymptote:
To find the vertical asymptotes, we need to check for the values of x that make the denominator of the function equal to zero. In this case, the denominator is (x^2 - 9). We can factorize it as (x - 3)(x + 3). So, the denominator is zero when x = -3 and x = 3.

Hence, the vertical asymptotes of f(x) occur at x = -3 and x = 3.

2. Horizontal Asymptote:
To find the horizontal asymptote, we will consider the degree of the numerator and denominator of the function. The degree of the numerator is 3, and the degree of the denominator is 2.

If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. The function approaches positive or negative infinity as x approaches positive or negative infinity.

If the degree of the numerator is equal to the degree of the denominator, then the horizontal asymptote is given by the ratio of the leading coefficients of the numerator and denominator.

In this case, since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

3. Slant Asymptote:
To find the slant or oblique asymptote, we can divide the numerator (2x^3 - 5x^2 - 19x + 1) by the denominator (x^2 - 9) using long division or synthetic division.

Performing long division, we get:

2x + 1
___________________________
x^2 - 9 | 2x^3 - 5x^2 - 19x + 1

2x^3 - 0x^2 - 18x
___________________
- x + 1

The quotient is 2x + 1 and the remainder is -x + 1.

Hence, the slant asymptote is given by the quotient, which is y = 2x + 1.

In summary:
- The vertical asymptotes are x = -3 and x = 3.
- There is no horizontal asymptote.
- The slant asymptote is y = 2x + 1.