# How do you find the slant asymptote of f(x) =( x^2 + 3x - 3) / (x+4)?

Jan 15, 2017

The slant asymptote is $y = x - 1$

#### Explanation:

Let's do the long division

$\textcolor{w h i t e}{a a a a}$${x}^{2} + 3 x - 3$$\textcolor{w h i t e}{a a a a}$∣$\textcolor{red}{x + 4}$

$\textcolor{w h i t e}{a a a a}$${x}^{2} + 4 x$$\textcolor{w h i t e}{a a a a a a a a}$∣$\textcolor{b l u e}{x - 1}$

$\textcolor{w h i t e}{a a a a a}$$0 - x - 3$

$\textcolor{w h i t e}{a a a a a a a}$$- x - 4$

$\textcolor{w h i t e}{a a a a a a a a a}$$0 + 1$

Therefore,

$\frac{{x}^{2} + 3 x - 3}{x + 4} = x - 1 + \frac{1}{x + 4}$

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

So,

$f \left(x\right) = x - 1 + \frac{1}{x + 4}$

${\lim}_{x \to - \infty} f \left(x\right) - \left(x - 1\right) = {\lim}_{x \to - \infty} \frac{1}{x} = {0}^{-}$

${\lim}_{x \to + \infty} f \left(x\right) - \left(x - 1\right) = {\lim}_{x \to + \infty} \frac{1}{x} = {0}^{+}$

The slant asymptote is $y = x - 1$

graph{(y-(x^2+3x-3)/(x+4))(y-x+1)=0 [-10, 10, -5, 5]}

Jan 15, 2017

Slant asymptote: $y = x - 1$. See the asymptotes-inclusive graph.

#### Explanation:

By division,

$y = f \left(x\right) = x - 1 + \frac{1}{x + 4}$

Rearranged,

$\left(y - x + 1\right) \left(x + 4\right) = 1$, revealing that the graph is a hyperbola with

asymptotes

$\left(y - x + 1\right) \left(x + 4\right) = 0$.

Separately,

y-x+1=0 gives a slant asymptote and

x+4=0 gives a vertical asymptote.

graph{((y-x+1)(x+4)-1)(y-x+1)(x+0.000001y+4)=0 [-23, 23.02, -12.7, 12.6]}