Question

How do you find vertical, horizontal and oblique asymptotes for `f(x) = (7x+1)/(2x-9)`?

Answer 1

Vertical asymptote is at `x = 4.5` and horizontal asymptote at
`y =3.5`

Explanation:

`f(x) = (7x+1)/(2x-9)`

For vertical asymptote denominator should be zero ,

`2x-9=0 :. x =9/2 =4.5`. So vertical asymptote is at `x = 4.5`

Here the degree of both denominator and numerator are same

then we have a horizontal asymptote at y = (numerator's leading

coefficient) / (denominator's leading coefficient) i.e at `y= 7/2` or

`y =3.5`. and no oblique asymptote.

Vertical asymptote is at `x = 4.5` and

horizontal asymptote at `y =3.5`

graph{(7x+1)/(2x-9) [-80, 80, -40, 40]}
[Ans]

Answer 2

`"vertical asymptote at "x=9/2`
`"horizontal asymptote at "y=7/2`

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 value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

`"solve " 2x-9=0rArrx=9/2" is the asymptote"`

`"horizontal asymptotes occur as"`

`lim_(xto+-oo),f(x)toc" ( a constant)"`

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

`f(x)=((7x)/x+1/x)/((2x)/x-9/x)=(7+1/x)/(2-9/x)`

as `xto+-oo,f(x)to(7+0)/(2-0)`

`rArry=7/2" is the asymptote"`

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no oblique asymptotes.
graph{(7x+1)/(2x-9) [-20, 20, -10, 10]}