How do you find the vertical, horizontal or slant asymptotes for #(x-7)/(3x^2+17x-6 )#?

1 Answer
Mar 5, 2017

Answer:

#"vertical asymptotes at "x=-6" and "x=1/3#
#"horizontal asymptote at "y=0#

Explanation:

#"let "f(x)=(x-7)/(x^2+17x-6)#

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

#"solve: "3x^2+17x-6=0rArr(3x-1)(x+6)=0#

#rArrx=-6" and "x=1/3" are the asymptotes"#

Horizontal asymptotes occur as

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

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

#f(x)=(x/x^2-7/x^2)/((3x^2)/x^2+(17x)/x^2-6/x^2)=(1/x-7/x^2)/(3+(17)/x-6/x^2)#

as #xto+-oo,f(x)to(0-0)/(3+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 1, denominator-degree 2 ) Hence there are no slant asymptotes.
graph{(x-7)/(3x^2+17x-6) [-10, 10, -5, 5]}