asymptotes

**samtrix** · October 3rd 2007, 11:32 PM

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!

**topsquark** · October 4th 2007, 04:11 AM

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!

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

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

So, for example, the possible rational zeros of the numerator are $\pm 1, \pm 2$ . I find that both x = -1 and x = -2 are zeros of the polynomial. Thus
$x^3 + 4x^2 + 5x + 2 = (x + 1)(x + 2)(x + p)$
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:
$x^3 + 4x^2 + 5x + 2 = (x + 1)^2(x + 2)$

You can do a similar factorization of the denominator.

-Dan

**Soroban** · October 4th 2007, 04:38 AM

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.

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

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

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

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

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

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

Vertical asymptotes: Value for which the denominator equals zero.

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

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

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

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

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

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

**samtrix** · October 4th 2007, 06:39 AM

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

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

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