1. ## asymptotes

hi guys.. need help in sketching graph and labeling intercepts and asymptotes of the attached function.

it has a power 3 x in the denominator .. how can i solve and simplify it?

thanks!

2. Originally Posted by samtrix
hi guys.. need help in sketching graph and labeling intercepts and asymptotes of the attached function.

it has a power 3 x in the denominator .. how can i solve and simplify it?

thanks!
$\displaystyle y = \frac{x^3 + 4x^2 + 5x + 2}{x^3 -x^2 - 8x + 12}$

We may use the rational root theorem: The rational zeros of a polynomial
$\displaystyle ax^n + bx^{n - 1} + ~ ... ~ + cx + d$
(if they exist) will be of the form:
$\displaystyle \pm \frac{\text{factor of d}}{\text{factor of a}}$

So, for example, the possible rational zeros of the numerator are $\displaystyle \pm 1, \pm 2$. I find that both x = -1 and x = -2 are zeros of the polynomial. Thus
$\displaystyle x^3 + 4x^2 + 5x + 2 = (x + 1)(x + 2)(x + p)$
We can find the remaining zero by long division, synthetic division, or by multiplying the above factorized expression out and solving for p. I get:
$\displaystyle x^3 + 4x^2 + 5x + 2 = (x + 1)^2(x + 2)$

You can do a similar factorization of the denominator.

-Dan

3. Hello, samtrix!

You should be familiar with the Factor Theorem and Remainder Theorem.
(If you're not, I'll let someone else explain them.)

Sketch the graph and label intercepts and asymptotes.

$\displaystyle y \:=\:\frac{x^3+4x^2 + 5x + 2}{x^3 - x^2 - 8x + 12}$

y-intercept: Let $\displaystyle x = 0$ and solve for $\displaystyle y.$

.$\displaystyle y \:=\:\frac{0^3 + 4\!\cdot\!0^2+5\!\cdot\!9 + 2}{0^3-0^2-8\!\cdot\!0 + 12} \:=\:\frac{2}{12} \:=\:\frac{1}{6}\quad\Rightarrow\quad\boxed{ \text{y-intercept: }\left(0,\,\frac{1}{6}\right)}$

x-intercepts: Let $\displaystyle y = 0$ and solve for $\displaystyle x.$

. . $\displaystyle x^3 + 4x^2 + 5x + 2\:=\:0\quad\Rightarrow\quad(x+1)^2(x+2) \:=\:0$

Therefore: .$\displaystyle x \:=\:-1,\,-2\quad\Rightarrow\quad\boxed{\text{x-intercepts: }(-1,\,0),\;(-2,\,0)}$

Vertical asymptotes: Value for which the denominator equals zero.

. . $\displaystyle x^3-x^2-8x + 12\:=\:0\quad\Rightarrow\quad(x-2)^2(x+3)\:=\:0$

Therefore, $\displaystyle \boxed{\text{vertical asymptotes: }x \,=\,2,\;x\,=\,-3}$

Horizontal asymptote: Find $\displaystyle \lim_{x\to\infty}y$

We have: .$\displaystyle \lim_{x\to\infty}\frac{x^3+4x^2+5x+2}{x^3-x^2-8x+12}$

Divide top and bottom by $\displaystyle x^3$
. . $\displaystyle \lim_{x\to\infty}\frac{\frac{x^3}{x^3} + \frac{4x^2}{x^3} + \frac{5x}{x^3} + \frac{2}{x^3}}{\frac{x^3}{x^3} - \frac{x^2}{x^3} - \frac{8x}{x^3} + \frac{12}{x^3}} \;=\;\lim_{x\to\infty}\frac{1 + \frac{4}{x} + \frac{5}{x^2} + \frac{2}{x^3}}{1 - \frac{1}{x} - \frac{8}{x^2} + \frac{12}{x}}$ . $\displaystyle =\;\frac{1 + 0 + 0 + 0}{1 - 0 - 0 + 0} \;=\;1$

Therefore: .$\displaystyle \boxed{\text{Horizontal asymptote: }y \:=\:1}$

4. thanks soroban and topsquark! i got the idea ^__^

and i will look up on factor theorem and remainder theoram.