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)

2 Answers
Mar 6, 2018

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

Mar 6, 2018

"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]}