f(x)=x^3-3x^2-3x-8/(-3x^2-4x-6)

Find the equation of the non-vertical asymptote.
What is the smallest value of x at which f(x) intersects its non-vertical asymptote?

please help i have no idea how to solve this thanks!

To find the equation of the non-vertical asymptote of the function f(x), you can follow these steps:

Step 1: First, we need to determine if the degree of the numerator (which is 3, the highest power of x) is equal to, less than, or greater than the degree of the denominator (which is 2, the highest power of x). In this case, the degree of the numerator is 3 and the degree of the denominator is 2.

Step 2: Since the degree of the numerator is higher than the degree of the denominator, the non-vertical asymptote will be a slant asymptote. To find the equation of the slant asymptote, we need to perform long division between the numerator and the denominator.

Performing the long division, we have:
___________________
-3x^2 - 4x - 6 | x^3 - 3x^2 - 3x - 8
- (x^3 + (4/3)x^2 + (2/3)x)
_____________________
- (7/3)x^2 - (11/3)x - 8
+ ((7/3)x^2 + (28/9)x + (14/9))
_____________________
- (25/9)x - (62/9)

Therefore, the equation of the slant asymptote is y = -25/9x - 62/9.

To find the smallest value of x at which f(x) intersects its non-vertical asymptote, we need to find the value of x where f(x) - (-25/9x - 62/9) = 0.

Simplifying the expression, we have:
(x^3 - 3x^2 - 3x - 8) - (-25/9x - 62/9) = 0
x^3 - 3x^2 - 3x - 8 + (25/9x + 62/9) = 0
x^3 - 3x^2 - 3x + 25/9x + 62/9 - 8 = 0
x^3 - 3x^2 + (16/9)x + 46/9 = 0

Unfortunately, finding the exact value of x at which f(x) intersects its non-vertical asymptote might be difficult algebraically. You can try solving it using numerical methods such as graphing the function or using a graphing calculator to find the approximate value.