Question

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

Answer 1

The vertical asymptotes are `x=2` and `x=-2`
The horizontal asymptote is `y=3`
No oblique asymptote

Explanation:

Let's factorise the numerator

`3x^2+2x-1=(3x-1)(x+1)`

The denominator is

`x^2-4=(x+2)(x-2)`

Therefore,

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

The domain of `f(x)` is `RR-{2,-2}`

To find the vertical asymptotes, we calculate

`lim_(x->2^-)f(x)=15/(0^-) = -oo`

`lim_(x->2^+)f(x)=15/(0^+) = +oo`

so,

The vertical asymptote is `x=2`

`lim_(x->-2^-)f(x)=7/(0^+) = +oo`

`lim_(x->-2^+)f(x)=7/(0^-) = -oo`

The vertical asymptote is `x=-2`

To calculate the horizontal asymptotes , we calculate the limit as `x->+-oo`

`lim_(x->+oo)f(x)=lim_(x->+oo)(3x^2)/(x^2)=3`

`lim_(x->-oo)f(x)=lim_(x->-oo)(3x^2)/(x^2)=3`

The horizontal asymptote is `y=3`

There is no oblique asymptote as thr degree of the numerator is `=` to the degree of the denominator

graph{(3x^2+2x-1)/(x^2-4) [-14.24, 14.24, -7.12, 7.12]}

Answer 2

`"vertical asymptotes at " x=+-2`
`"horizontal asymptote at "y=3`

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 "x^2-4=0rArr(x-2)(x+2)=0`

`rArrx=-2" and " x=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)=((3x^2)/x^2+(2x)/x^2-1/x^2)/(x^2/x^2-4/x^2)=(3+2/x-1/x^2)/(1-4/x^2)`

as `xto+-oo,f(x)to(3+0-0)/(1-0)`

`rArry=3" is the asymptote"`

`"there are no removable discontinuities"`
graph{(3x^2+2x-1)/(x^2-4) [-10, 10, -5, 5]}