Question

How do you find the asymptotes for `y = (x^2 + 2x - 3)/( x^2 - 5x - 6)`?

Answer 1

`"vertical asymptotes at "x=-1" and "x=6`
`"horizontal asymptote at "y=1`

Explanation:

The denominator of y cannot be zero as this would make y 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 "x^2-5x-6=0rArr(x-6)(x+1)=0`

`x=-1" and "x=6" are the asymptotes"`

`"Horizontal asymptotes occur as"`

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

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

`y=(x^2/x^2+(2x)/x^2-3/x^2)/(x^2/x^2-(5x)/x^2-6/x^2)=(1+2/x-3/x^2)/(1-5/x-6/x^2)`

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

`y=1" is the asymptote"`
graph{(x^2+2x-3)/(x^2-5x-6) [-20, 20, -10, 10]}