How do you find vertical, horizontal and oblique asymptotes for f(x)=( 21 x^2 ) / ( 3 x + 7)?

Dec 6, 2016

The vertical asymptote is $x = - \frac{7}{3}$
The oblique asymptote is $y = 7 x - \frac{49}{3}$
No horizontal asymptote.

Explanation:

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- \frac{7}{3}\right\}$

The denominator must be $\ne 0$, $\implies$, $x \ne - \frac{7}{3}$

So, the vertical asymptote is $x = - \frac{7}{3}$

As the degree of the numerator is $>$ then the degree of the denominator, we expect an oblique asymptote.

Let's do a long division

$\textcolor{w h i t e}{a a a a}$$21 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a}$∣$3 x + 7$

$\textcolor{w h i t e}{a a a a}$$21 {x}^{2} + 49 x$$\textcolor{w h i t e}{a a a a a a}$∣$7 x - \frac{49}{3}$

$\textcolor{w h i t e}{a a a a a a a}$$0 - 49 x$

$\textcolor{w h i t e}{a a a a a a a a a}$$- 49 x - \frac{343}{3}$

$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$0 + \frac{343}{3}$

So,

$\frac{21 {x}^{2}}{3 x + 7} = 7 x - \frac{49}{3} + \frac{\frac{343}{3}}{3 x + 7}$

The oblique asymptote is $y = 7 x - \frac{49}{3}$

To calculate the limits of $x \to \pm \infty$, you take the terms of highest degree in the numerator and the denominator

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} \frac{21 {x}^{2}}{3 x} = {\lim}_{x \to \pm \infty} 7 x = \pm \infty$

So, there is no horizontal asymptote.

graph{(y-(21x^2)/(3x+7))(y-7x+49/3)=0 [-139.1, 127.9, -106, 27.5]}