Question

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

Answer 1

`"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]}