Question

What are the horizontal and vertical asymptotic (if any) of the curve f(x)=(12x+52)/(3x^2+2x-1)?

Find the horizontal and vertical asymptotic if any of the curve `f(x)=(12x+52)/(3x^2+2x-1)`

Answer 1

Verticsl asymptote at `x=1/3 and x=-1`
Horizontal asymptote `y=0`

Explanation:

You can find the vertical asymptote if you equate the denominator to 0

`3x^2 +2x -1=0`
`(3x-1)(x+1)=0`
`x=1/3 , x=-1`

Horizontal asymptote is at y=0 as the highest degree ( power on the x) on the numerator is less than the highest degree on the denominator

Answer 2

`"vertical asymptotes at "x=-1" and "x=1/3`
`"horizontal asymptote at "y=0`

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 "3x^2+2x-1=0rArr(3x-1)(x+1)=0`

`rArrx=-1" and "x=1/3" 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)=((12x)/x^2+52/x^2)/((3x^2)/x^2+(2x)/x^2-1/x^2)=(12/x+52/x^2)/(3+2/x-1/x^2)`

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

`rArry=0" is the asymptote"`
graph{(12x+52)/(3x^2+2x-1) [-20, 20, -10, 10]}