Answer
The vertical asymptotes: $x=-4$ and $x=\frac{2}{3}$.
The horizontal asymptote: $y=-\frac{8}{3}$.
Work Step by Step
We have:
$$f(x)=\frac{-8x^2+10x+3}{3x^2+10x-8}=\frac{-(4x+1)(2x-3)}{(3x-2)(x+4)}$$
Finding the vertical asymptotes, equate the denominator to $0$:
$$3x^2+10x-8=0$$ $$(3x-2)(x+4)=0$$ $$3x-2=0$$ $$3x=2$$ $$x=\frac{2}{3}$$ $$x+4=0$$ $$x=-4$$
Thus, the vertical asymptotes are $x=-4$ and $x=\frac{2}{3}$.
Since the degree of the numerator and the degree of the denominator are equal, taking the quotient of the leading coefficient of the numerator and the leading coefficient of the denominator, the horizontal asymptote is:
$$y=\frac{-8}{3}$$ $$y=-\frac{8}{3}$$
Thus, the horizontal asymptote is $y=-\frac{8}{3}$.
