Question

What are the asymptotes and removable discontinuities, if any, of `f(x)=(4(x+2)(x-1))/(3(x+2)(x-5)`?

Answer 1

`"vertical asymptote at " x=5`
`"horizontal asymptote at " y=4/3`
`"removable discontinuity at " (-2,4/7)`

Explanation:

`"simplify f(x) by cancelling common factors"`

`f(x)=(4cancel((x+2))(x-1))/(3cancel((x+2))(x-5))=(4(x-1))/(3(x-5))`

Since we have removed the factor ( x + 2 ) there will be a removable discontinuity at x = - 2 (hole )

`f(-2)=(4(-3))/(3(-7))=(-12)/(-21)=4/7`

`rArr" point discontinuity at " (-2,4/7)`

The graph of `f(x)=(4(x-1))/(3(x-5)) "will be the same as"`

`(4(x+2)(x-1))/(3(x+2)(x-5))" but without the hole"`

The denominator of f(x) cannot be zero as this would make f(x) 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 " 3(x-5)=0rArrx=5" is the asymptote"`

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

`f(x)=((4x)/x-4/x)/((3x)/x-15/x)=(4-4/x)/(3-15/x)`

as `xto+-oo,f(x)to(4-0)/(3-0`

`rArry=4/3" is the asymptote"`
graph{(4x-4)/(3x-15) [-16.02, 16.01, -8.01, 8.01]}