# How do you find vertical, horizontal and oblique asymptotes for (3x^2+2x-5)/(x-4)?

##### 1 Answer
Jan 12, 2017

The vertical asymptote is $x = 4$
The oblique asymptote is $y = 3 x + 14$
No horizontal asymptote

#### Explanation:

Let $f \left(x\right) = \frac{3 {x}^{2} + 2 x - 5}{x - 4}$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{4\right\}$

As we cannot divide by $0$, $\implies$, $x \ne 4$

The vertical asymptote is $x = 4$

As the degree of the numerator is $>$ than the degree of the denominator, there is an oblique asymptote.

Let's do a long division

$\textcolor{w h i t e}{a a a a}$$3 {x}^{2} + 2 x - 5$$\textcolor{w h i t e}{a a a a a a}$∣$\textcolor{b l u e}{x - 4}$

$\textcolor{w h i t e}{a a a a}$$3 {x}^{2} - 12 x$$\textcolor{w h i t e}{a a a a a a a a a}$∣$\textcolor{red}{3 x + 14}$

$\textcolor{w h i t e}{a a a a a a}$$0 + 14 x - 5$

$\textcolor{w h i t e}{a a a a a a a a}$$+ 14 x - 56$

$\textcolor{w h i t e}{a a a a a a a a a a}$$+ 0 + 51$

Therefore,

$f \left(x\right) = \left(3 x + 14\right) + \frac{51}{x - 4}$

Now, we calculate the limits

${\lim}_{x \to - \infty} \left(f \left(x\right) - \left(3 x + 14\right)\right) = {\lim}_{x \to - \infty} \frac{51}{x} = {0}^{-}$

${\lim}_{x \to + \infty} \left(f \left(x\right) - \left(3 x + 14\right)\right) = {\lim}_{x \to + \infty} \frac{51}{x} = {0}^{+}$

The oblique asymptote is $y = 3 x + 14$

graph{(y-(3x^2+2x-5)/(x-4))(y-3x-14)=0 [-106, 160.9, -31.5, 102.1]}