Question

How do you find the asymptotes for `(9x^2 – 36) /( x^2 - 9)`?

Answer 1

`"vertical asymptotes at "x=+-3`
`"horizontal asymptote at "y=9`

Explanation:

`"let "f(x)=(9x^2-36)/(x^2-9)`

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 "x^2-9=0rArr(x-3)(x+3)=0`

`x=+-3" 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)=((9x^2)/x^2-(36)/x^2)/(x^2/x^2-9/x^2)=(9-(36)/x^2)/(1-9/x^2`

`"as "xto+-oo,f(x)to(9-0)/(1-0)`

`y=9" is the asymptote"`
graph{(9x^2-36)/(x^2-9) [-40, 40, -20, 20]}

Answer 2

vertical asymptotes: `x=\pm 3`

horizontal asymptote: `y=9`

Explanation:

The given function:

`f(x)=\frac{9x^2-36}{x^2-9}`

Vertical asymptotes:

The above function will have vertical asymptote where denominator becomes zero i.e.

`\therefore x^2-9=0`

`x^2=9`

`x=\pm 3`

Horizontal asymptotes:

The above function will have horizontal where

`y=\lim_{x\to \pm \infty}\frac{9x^2-36}{x^2-9}`

`y=\lim_{x\to \pm \infty}\frac{9-36/x^2}{1-9/x^2}`

`y=\frac{9-0}{1-0}`

`y=9`

Hence, vertical asymptotes: `x=\pm 3`

horizontal asymptote: `y=9`