Question

What is the horizontal and vertical asumptotes of `f(x) = (7x^2)/(9x^2-16)`?

Answer 1

`"vertical asymptotes at" x=+-4/3`
`"horizontal asymptote at " y=7/9`

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: `9x^2-16=0rArrx^2=16/9rArrx=+-4/3`

`rArrx=-4/3" and " x=4/3" 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)=((7x^2)/x^2)/((9x^2)/x^2-16/x^2)=7/(9-16/x^2)`

as `xto+-oo,f(x)to7/(9-0)`

`rArry=7/9" is the asymptote"`
graph{(7x^2)/(9x^2-16) [-10, 10, -5, 5]}

Answer 2

The vertical asymptotes are `x=-4/3` and `x=4/3`
The horizontal asymptote is `y=7/9`

Explanation:

The denominator
x
`=9x^2-16=(3x-4)(3x+4)`

The domain of `f(x)` is `D_f(x)=RR-{-4/3,4/3}`

As we cannot divide by `0`, `x!=-4/3` and `x!=4/3`

The vertical asymptotes are `x=-4/3` and `x=4/3`

To find the horizontal limits, we calculate the limits of `f(x)` as `x->+-oo`

We take the terms of highest degree in the numerator and the denominator.

x`lim_(x->+-oo)f(x)=lim_(x->+-oo)(7x^2)/(9x^2)=7/9`

The horizontal asymptote is `y=7/9`
graph{7x^2/(9x^2-16) [-10, 10, -5, 5]}