How do you find the vertical, horizontal or slant asymptotes for #(4x)/(x^2-25)#?

1 Answer
Jan 7, 2017

Answer:

#"vertical asymptotes at " x=+-5#

#"horizontal asymptote at " y=0#

Explanation:

The denominator of the function cannot be zero as this would make the function 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 : #x^2-25=0rArrx^2=25rArrx=+-5#

#rArrx=-5" and " x=5" 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)=((4x)/x^2)/(x^2/x^2-25/x^2)=(4/x)/(1-25/x^2)#

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

#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{(4x)/(x^2-25) [-10, 10, -5, 5]}