# Finding equation of asymptotes-Specialists maths

• Dec 27th 2010, 11:21 PM
johnsy123
Finding equation of asymptotes-Specialists maths
The equations of the asymptotes of the curve with the equation-

y=4x+3\x^2-9
• Dec 27th 2010, 11:27 PM
earboth
Quote:

Originally Posted by johnsy123
The equations of the asymptotes of the curve with the equation-

y=4x+3/x^2-9

1. Re-arrange the given equation:

$\displaystyle y = \dfrac{4x+3}{x^2-9} = \dfrac{4x+3}{(x-3)(x+3)}$

2. The zeros of the denominator will yield the equations of the vertical asymptotes.

3. Re-arrange the given equation by long division:

$\displaystyle (4x+3)\div (x^2-9)=\underbrace{0}_{\text{term of asymptote}}+\dfrac{4x+3}{x^2-9}$

That means you'll get a horizontal asymptote since the term of the asymptote is a constant (the slope is zero)
• Dec 27th 2010, 11:32 PM
johnsy123
Sorry i made an error. The equation is y= 4x +3/x^2-9.
• Dec 27th 2010, 11:36 PM
FernandoRevilla
These problems are just routine if you know the corresponding theory. For example:

(a) Horizontal:

Find $\displaystyle a=\displaystyle\lim_{x \to \infty}{f(x)}$

(b) Vertical:

Find finite values of $\displaystyle x_0$ such that $\displaystyle y\rightarrow{\infty}$ as $\displaystyle x\rightarrow x_0$

(iii) Oblique

...

Try it.

Fernando Revilla
• Dec 27th 2010, 11:50 PM
johnsy123
So what your simplistically saying, is to find the horizontal asymptote i should sub in infinite x values and to find the vertical asymptote i should sub in infinite y values?
• Dec 28th 2010, 12:22 AM
Prove It
Quote:

Originally Posted by earboth
1. Re-arrange the given equation:

$\displaystyle y = \dfrac{4x+3}{x^2-9} = \dfrac{4x+3}{(x-3)(x+3)}$

2. The zeros of the denominator will yield the equations of the vertical asymptotes.

3. Re-arrange the given equation by long division:

$\displaystyle (4x+3)\div (x^2-9)=\underbrace{0}_{\text{term of asymptote}}+\dfrac{4x+3}{x^2-9}$

That means you'll get a horizontal asymptote since the term of the asymptote is a constant (the slope is zero)

Better yet, long divide first :) Then your constant will be the horizontal asymptote and the zeros of the denominator will be your vertical (or other) asymptotes.

And no johnsy, you can't substitute $\displaystyle \displaystyle \infty$ because $\displaystyle \displaystyle \infty$ is not a number. But you can think about what would happen as your function gets infinitely large in either direction...