Answer
$(\frac{7\pm\sqrt {241}}{2}, 1)$
Work Step by Step
Step 1. We can identify its horizontal asymptote as $y=1$ (ratio of leading coefficients).
Step 2. Let $f(x)=1$, we have $-3x^3-21x^2+43x+60=-6x^3+x^2+24x-20$ or $3x^3-22x^2+19x+80=0$
Step 3. Use synthetic division as shown in the figure, $x=5$ is a zero of the above equation but not a point on the curve (because a hole).
Step 4. Use the quotient for the other zeros: $3x^2-7x-16=0$ which gives $x=\frac{7\pm\sqrt {7^2+4(3)(16)}}{2}=\frac{7\pm\sqrt {241}}{2}$
Step 5. Thus the intersects are $(\frac{7\pm\sqrt {241}}{2}, 1)$
