# How do you find the vertical, horizontal or slant asymptotes for #y=(2x^2 + 3)/(x^2 - 6 )#?

##### 1 Answer

Mar 15, 2016

vertical asymptotes

horizontal asymptote y = 2

#### Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s equate the denominator to zero.

solve :

# x^2 - 6 = 0 → x^2 = 6 → x = ± sqrt6# Horizontal asymptotes occur as

#lim_(x→±∞) f(x) → 0# divide all terms on numerator / denominator by

# x^2 #

#(2x^2+3)/(x^2-6) =( (2x^2)/x^2 + 3/x^2)/(x^2/x^2 - 6/x^2)=(2+3/x^2)/(1-6/x^2)# now as x →∞

# , 3/x^2 " and " 6/x^2 → 0 #

# rArr y = 2 " is the asymptote "# Slant asymptotes occur when the degree of the numerator is greater than the degree of the denominator.

This is not the case here hence no slant asymptotes.

Here is the graph of the function.

graph{(2x^2+3)/(x^2-6) [-10, 10, -5, 5]}