Question

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

Answer 1

`"vertical asymptotes at "x=3/2" and "x=7`

`"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)(x-7)=0`

`rArrx=3/2" and " x=7" are the asymptotes"`

Horizontal asymptotes occur as

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

`f(x)=-x/(2x^2-17x+21)`

divide terms on numerator/denominator by the highest power of x, that is `x^2`

`f(x)=-(x/x^2)/((2x^2)/x^2-(17x)/x^2+(21)/x^2)=-(1/x)/(2-17/x+(21)/x^2`

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

`rArry=0" is the asymptote"`
graph{-(x)/((2x-3)(x-7)) [-10, 10, -5, 5]}