I know I am looking to find a slanted horizontal asymptote because I can see $6x^3$ is in the numerator and is one degree more than $3x^2$. The method, in the book, to find the equation of the slanted asymptote is to use synthetic division. Not sure how to use the book's way because I keep finding imaginary numbers for factors. Since I am looking at only real numbers in the domain, I feel like I am really lost.
When I evaluate the denominator of $\frac{6x^3-5x}{3x^2+4}$ I get: $$3x^2+4=0$$ $$3x^2=-4$$$$x^2=-\frac43$$$$x=\pm\sqrt{-\frac43}$$$$x=\pm\frac{2i\sqrt3}{3}$$
I don't think I can synthetic divide with imaginary numbers, also I don't think these numbers are in the domain of the function. I really am wondering what I can do to find the asymptote? thanks
Indeed the function is defined for all real $x$, as you have shown. Take the limit to infinity to find the asymptote:
$$\lim_{x \to \infty} \frac{6x^3-5x}{3x^2+4} = \lim_{x \to \infty} \frac{6x-5/x}{3+4/x^2} = \ ?$$
Without calculus, you can write:
$$6x^3 - 5x = 2x(3x^2 + 4) - 13x$$
and now divide by $3x^2+4$ on both sides.