# Thread: Finding equation of asymptotes-Specialists maths

1. ## Finding equation of asymptotes-Specialists maths

The equations of the asymptotes of the curve with the equation-

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

2. 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:

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

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

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

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

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

3. Sorry i made an error. The equation is y= 4x +3/x^2-9.

4. These problems are just routine if you know the corresponding theory. For example:

(a) Horizontal:

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

(b) Vertical:

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

(iii) Oblique

...

Try it.

Fernando Revilla

5. 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?

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

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

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

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

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

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 $\displaystyle \infty$ because $\displaystyle \infty$ is not a number. But you can think about what would happen as your function gets infinitely large in either direction...