How do you find vertical, horizontal and oblique asymptotes for #f(x) = (4x)/(x^2-25)#?

1 Answer
Jun 28, 2016

Answer:

vertical asymptotes x = ± 5
horizontal asymptote y = 0

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve: #x^2-25=0rArr(x-5)(x+5)=0rArrx=±5#

#rArrx=-5,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#

#((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"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1 and denominator-degree 2 ) Hence there are no oblique asymptotes.
graph{(4x)/(x^2-25) [-10, 10, -5, 5]}