Question

What are the asymptotes and removable discontinuities, if any, of `f(x)=(x-2)/(2x^2+5x)`?

Answer 1

`"vertical asymptotes at " x=0" and " x=-5/2`
`"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^2+5x=0rArrx(2x+5)=0`

`rArrx=0" and " x=-5/2" 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/x^2-2/x^2)/((2x^2)/x^2+(5x)/x^2)=(1/x-2/x^2)/(2+5/x)`

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

`rArr"asymptote is " y=0`
graph{(x-2)/(2x^2+5x) [-10, 10, -5, 5]}