Question

How do you find the vertical, horizontal and slant asymptotes of: `y = (1-5x)/(1+2x)`?

Answer 1

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

Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator 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=0rArrx=-1/2" is the asymptote"`

Horizontal asymptotes occur as

`lim_(xto+-oo),ytoc" (a constant)"`

divide terms on numerator/denominator by x

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

as `xto+-oo,yto(0-5)/(0+2)`

`rArry=-5/2" is the asymptote"`

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ) Hence there are no slant asymptotes.
graph{(1-5x)/(1+2x) [-10, 10, -5, 5]}