Question

How do you find the asymptotes for `f(x)=(1-5x)/(1+2x)`?

Answer 1

vertical asymptote `x=-1/2`
horizontal asymptote `y=-5/2`

Explanation:

The denominator of f(x) cannot equal zero. This would give division by zero which is undefined. Setting the denominator equal to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: 1 + 2x =0 `rArrx=-1/2" is the asymptote"`

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

`(1/x-(5x)/x)/(1/x+(2x)/x)=(1/x-5)/(1/x+2)`

as `xto+-oo,f(x)to(0-5)/(0+2)`

`rArry=-5/2" is the asymptote"`
graph{(1-5x)/(1+2x) [-10, 10, -5, 5]}