How do you find the vertical, horizontal and slant asymptotes of: #f(x) =2/(x^2+3x-10)#?

1 Answer
Jul 5, 2016

Answer:

vertical asymptotes x = -5 , x = 2
horizontal asymptote y = 0

Explanation:

The denominator cannot be zero as this would yield division by zero which is undefined. Setting the denominator equal to zero and solving for x will give the values that x cannot be and if the numerator is non-zero for these x then they are vertical asymptotes.

solve: #x^2+3x-10=0rArr(x+5)(x-2)=0#

#rArrx=-5,x=2" are the asymptotes"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by the highest exponent of x. that is #x^2#

#(2/x^2)/(x^2/x^2+(3x)/x^2-10/x^2)=(2/x^2)/(1+3/x-10/x^2)#

as #xto+-oo,f(x)to0/(1+0-0)#

#rArry=0" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 0, denominator-degree 2) Hence there are no slant asymptotes.
graph{(2)/(x^2+3x-10) [-10, 10, -5, 5]}