# How do you find the vertical, horizontal or slant asymptotes for #F(x)= (X^2-16)/(2X^2+3X-9)#?

##### 1 Answer

Mar 10, 2016

vertical asymptotes

horizontal asymptote

#### Explanation:

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

solve:

#2x^2+3x-9 = 0 → (2x-3)(x+3) = 0 #

#rArr x = -3 , x = 3/2 " are the asymptotes " # Horizontal asymptotes occur as

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

# x^2#

#(x^2-16)/(2x^2+3x-9) = (x^2/x^2-16/x^2)/((2x^2)/x^2+(3x)/x^2-9/x^2#

# = (1-16/x^2)/(2+3/x-9/x^2) # as x →∞ ,

#16/x^2 , 3/x " and " 9/x^2 → 0 #

#rArr y = 1/2 " is the asymptote "# Here is the graph of the function.

graph{(x^2-16)/(2x^2+3x-9) [-10, 10, -5, 5]}