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

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Find 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

##### 2 Answers

Verticsl asymptote at

Horizontal asymptote

#### Explanation:

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

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

#### 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]}