Question

What are the asymptotes of `f(x)=-x/((2x-3)(7x-1))`?

Answer 1

vertical asymptotes at `x=1/7" and "x=3/2`
horizontal asymptote at y = 0

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) 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 : `(2x-3)(7x-1)=0`

`rArrx=1/7" and "x=3/2" 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)=-x/(14x^2-23x+3)=-(x/x^2)/((14x^2)/x^2-(23x)/x^2+3/x^2)`

`=-(1/x)/(14-23/x+3/x^2)`

as `xto+-oo,f(x)to0/(14-0+0)`

`rArry=0" is the asymptote"`
graph{-(x)/((2x-3)(x-1)) [-20, 20, -10, 10]}