# 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 $f \left(x\right) = \frac{12 x + 52}{3 {x}^{2} + 2 x - 1}$

Mar 6, 2018

Verticsl asymptote at $x = \frac{1}{3} \mathmr{and} x = - 1$
Horizontal asymptote $y = 0$

#### Explanation:

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

$3 {x}^{2} + 2 x - 1 = 0$
$\left(3 x - 1\right) \left(x + 1\right) = 0$
$x = \frac{1}{3} , x = - 1$

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

$\text{vertical asymptotes at "x=-1" and } x = \frac{1}{3}$
$\text{horizontal asymptote at } y = 0$

#### Explanation:

$\text{the denominator of f(x) cannot be zero as this would}$
$\text{make f(x) undefined. Equating the denominator to }$
$\text{zero and solving gives the values that x cannot be and}$
$\text{if the numerator is non-zero for these values then they}$
$\text{are vertical asymptotes}$

$\text{solve } 3 {x}^{2} + 2 x - 1 = 0 \Rightarrow \left(3 x - 1\right) \left(x + 1\right) = 0$

$\Rightarrow x = - 1 \text{ and "x=1/3" are the asymptotes}$

$\text{horizontal asymptotes occur as}$

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{( a constant)}$

$\text{divide terms on numerator/denominator by the }$
$\text{highest power of x that is } {x}^{2}$

$f \left(x\right) = \frac{\frac{12 x}{x} ^ 2 + \frac{52}{x} ^ 2}{\frac{3 {x}^{2}}{x} ^ 2 + \frac{2 x}{x} ^ 2 - \frac{1}{x} ^ 2} = \frac{\frac{12}{x} + \frac{52}{x} ^ 2}{3 + \frac{2}{x} - \frac{1}{x} ^ 2}$

$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{0 + 0}{3 + 0 - 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{(12x+52)/(3x^2+2x-1) [-20, 20, -10, 10]}