How do you find the slant asymptote of f(x) = (3x^3 - 28x^2 +54x - 24) / (x^2 - 8x + 7)?

Dec 6, 2015

The slant asymptote is the line $y = 3 x - 4$ as $x \setminus \to \setminus \pm \setminus \infty$

Explanation:

If a function has a slant asymptote, it means that it is asymptotically equivalent to a certain line $y = m x + q$.

This means that

${\lim}_{x \setminus \to \setminus \infty} f \left(x\right) - \left(m x + q\right) = 0$.

Divide by $x$:

${\lim}_{x \setminus \to \setminus \infty} f \frac{x}{x} - \left(m + \frac{q}{x}\right) = 0$.

Since $\frac{q}{x} \setminus \to 0$, we have that

${\lim}_{x \setminus \to \setminus \infty} f \frac{x}{x} - m = 0$.

And so $m = {\lim}_{x \setminus \to \setminus \infty} f \frac{x}{x}$

Once $m$ is known, we can find $q$ by calculating

${\lim}_{x \setminus \to \setminus \infty} f \left(x\right) - m x = q$.

Now, let's do the calulations: in your case,

${\lim}_{x \setminus \to \setminus \pm \setminus \infty} f \frac{x}{x} = 3$, since we have

$f \frac{x}{x} = \setminus \frac{3 {x}^{3} - 28 {x}^{2} + 54 x - 24}{{x}^{3} - 8 {x}^{2} + 7 x}$

And the limits at $\setminus \pm \setminus \infty$ are given by the ratio of the coefficients of the leading terms (i.e. ${x}^{3}$), which in this case are $3$ and $1$.

So, $m = 3$. Now let's compute $q$:

$\setminus {\lim}_{x \setminus \to \setminus \pm \setminus \infty} f \left(x\right) - m x = \setminus \frac{3 {x}^{3} - 28 {x}^{2} + 54 x - 24}{{x}^{2} - 8 x + 7} - 3 x$

And since

$\setminus \frac{3 {x}^{3} - 28 {x}^{2} + 54 x - 24}{{x}^{2} - 8 x + 7} - 3 x = \setminus \frac{3 {x}^{3} - 28 {x}^{2} + 54 x - 24 - \left(3 {x}^{3} - 24 {x}^{2} + 21 x\right)}{{x}^{2} - 8 x + 7}$
$= \setminus \frac{- 4 {x}^{2} + 33 x - 24}{{x}^{2} - 8 x + 7}$

And this function tends to $- 4$ as $x \setminus \to \setminus \pm \setminus \infty$