Question

How do you find the asymptotes for `y=(x+3)/((x-4)(x+3))`?

Answer 1

`"vertical asymptote at "x=4`
`"horizontal asymptote at "y=0`

Explanation:

`"simplify by cancelling"`

`y=cancel(x+3)/((x-4)cancel((x+3)))=1/(x-4)`

`"the removal of "(x+3)" from numerator/denominator"`
`"indicates a hole at "x=-3`

`"the graph of the simplification "1/(x-4)" is the same as"`

`"the graph of "y=(x+3)/((x-4)(x+3))" without the hole"`

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 "x-4=0rArrx=4" is the asymptote"`

`"Horizontal asymptotes occur as"`

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

`"divide terms on numerator/denominator by "x`

`y=(1/x)/(x/x-4/x)=(1/x)/(1-4/x)`

`"as "xto+-oo,yto0/(1-0)`

`y=0" is the asymptote"`
graph{1/(x-4) [-10, 10, -5, 5]}