# Graph ((12n^2-27)/(2n^2-5n+3))÷((8n^2+10-3)/(n^2-6n+5))

12n2-272n2-5n+3÷8n2+10-3n2-6n+5
Find where the expression 3(x-5)(2x+3)8×2+7 is undefined.

The vertical asymptotes occur at areas of infinite discontinuity.
No Vertical Asymptotes
Consider the rational function R(x)=axnbxm where n is the degree of the numerator and m is the degree of the denominator.
1. If n<m, then the x-axis, y=0, is the horizontal asymptote.
2. If n=m, then the horizontal asymptote is the line y=ab.
3. If n>m, then there is no horizontal asymptote (there is an oblique asymptote).
Find n and m.
n=2
m=2
Since n=m, the horizontal asymptote is the line y=ab where a=6 and b=8.
y=34
There is no oblique asymptote because the degree of the numerator is less than or equal to the degree of the denominator.
No Oblique Asymptotes
This is the set of all asymptotes.
No Vertical Asymptotes
Horizontal Asymptotes: y=34
No Oblique Asymptotes
Graph ((12n^2-27)/(2n^2-5n+3))÷((8n^2+10-3)/(n^2-6n+5))