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

Answer:

vertical asymptotes # x = -3 , x = 3/2#
horizontal asymptote #y = 1/2#

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]}