Question

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

Answer 1

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]}