r(x)=3x^2-13x+12/x^2-5x+6. first factor, then find the horizontal and vertical asymptotes and the coordinates of the hole in the graph

Question

anonymous · Answer

I got it factored to (3x+4)(x+3)/(x-3)(x-2)

anonymous · Answer

the numerator is factored incorrectly

anonymous · Answer

the "middle term" is $-13x$ so it would factor as $(3x-4)(x-3)$

anonymous · Answer

ok I see!

anonymous · Answer

horizontal asymptote would be y=3 and the vertical ones would be x=3 and x=2?

anonymous · Answer

no

anonymous · Answer

$$\frac{(3x-4)(x-3)}{(x-3)(x-2)}$$ you can cancel the common factor of $x-3$ which means there is a "hole" at $x=3$ not a vertical asymptote

anonymous · Answer

that gives you 
$$\frac{3x-4}{x-2}, x\neq 3$$ so the only vertical asymptote is $x=2$

anonymous · Answer

your horizontal asymptote is correct, it is $y=3$

anonymous · Answer

ok awesome! thank you so whenever you can cancel out like that, is that the hole?