# Thread: Finding asymptotes to the curve.

1. ## Finding asymptotes to the curve.

Find all the Asymptotes to the curve

$(2x+3)y = (x-1)^2$

2. Originally Posted by varunnayudu
Find all the Asymptotes to the curve

$(2x+3)y = (x-1)^2$
$y = \frac{x^2 - 2x + 1}{2x + 3} = \frac{x}{2} - \frac{7}{4} + \frac{25}{4(2x + 3)}$.

The equations of the vertical asymptote and the oblique asymptote should now be clear .....

3. Originally Posted by mr fantastic
$y = \frac{x^2 - 2x + 1}{2x + 3} = \frac{x}{2} - \frac{7}{4} + \frac{25}{4(2x + 3)}$.

The equations of the vertical asymptote and the oblique asymptote should now be clear .....

hw did u get this ........

4. Originally Posted by zorro
hw did u get this ........
Divide 2x + 3 into x^2 - 2x + 1 using polynomial long division.

5. Originally Posted by varunnayudu
Find all the Asymptotes to the curve

$(2x+3)y = (x-1)^2$
1. The equation of the vertical asymptote $x = {\text{const}}$ of $y = \frac{{\left( {x - 1} \right)^2 }}{{2x + 3}}$:

$2x + 3 = 0{\text{ }} \Leftrightarrow {\text{ }}2x = - 3{\text{ }} \Leftrightarrow {\text{ }}x = - \frac{3}{2}$ is the vertical asymptote.

2. The equation of the oblique asymptote $y = ax + b$ of $y = \frac{{\left( {x - 1} \right)^2 }}{{2x + 3}}$:

$a = \mathop {\lim }\limits_{x \to \infty } \frac{{f\left( x \right)}}
{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {x - 1} \right)^2 }}
{{\left( {2x + 3} \right)x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{x^2 - 2x + 1}}{{2x^2 + 3x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{1 - \frac{2}{x} + \frac{1}{{x^2 }}}}{{2 + \frac{3}{x}}} = \frac{1}{2}.$

$b = \mathop {\lim }\limits_{x \to \infty } \left[ {f\left( x \right) - ax} \right] = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{\left( {x - 1} \right)^2 }}{{2x + 3}} - \frac{x}{2}} \right] = \mathop {\lim }\limits_{x \to \infty } \frac{{ - 7x + 2}}{{4x + 6}} = \mathop {\lim }\limits_{x \to \infty } \frac{{ - 7 + \frac{2}{x}}}{{4 + \frac{6}{x}}} = \frac{{ - 7}}{4}.$

The oblique asymptote is $y = \frac{x}{2} - \frac{7}{4}.$

6. Originally Posted by DeMath
1. The equations of the vertical asymptote $x = {\text{const}}$ of $y = \frac{{\left( {x - 1} \right)^2 }}{{2x + 3}}$:

$2x + 3 = 0{\text{ }} \Leftrightarrow {\text{ }}2x = - 3{\text{ }} \Leftrightarrow {\text{ }}x = - \frac{3}{2}$ is the vertical asymptote.

2. The equations of the oblique asymptote $y = ax + b$ of $y = \frac{{\left( {x - 1} \right)^2 }}{{2x + 3}}$:

$a = \mathop {\lim }\limits_{x \to \infty } \frac{{f\left( x \right)}}
{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {x - 1} \right)^2 }}
{{\left( {2x + 3} \right)x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{x^2 - 2x + 1}}{{2x^2 + 3x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{1 - \frac{2}{x} + \frac{1}{{x^2 }}}}{{2 + \frac{3}{x}}} = \frac{1}{2}.$

$b = \mathop {\lim }\limits_{x \to \infty } \left[ {f\left( x \right) - ax} \right] = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{\left( {x - 1} \right)^2 }}{{2x + 3}} - \frac{x}{2}} \right] = \mathop {\lim }\limits_{x \to \infty } \frac{{ - 7x + 2}}{{4x + 6}} = \mathop {\lim }\limits_{x \to \infty } \frac{{ - 7 + \frac{2}{x}}}{{4 + \frac{6}{x}}} = \frac{{ - 7}}{4}.$

The oblique asymptote is $y = \frac{x}{2} - \frac{7}{4}.$

I suspect that your method for #2 will confuse the OP. Dividing first is the best approach I think.

7. Sorry, what is OP?
Is it Originally Posted?

8. Originally Posted by DeMath
Sorry, what is OP?
Is it Originally Posted?
OP = Original Poster (the member who started the thread).

9. Originally Posted by mr fantastic
I suspect that your method for #2 will confuse the OP.
Why?

Dividing first is the best approach I think.
The method which I offered is an analytical method, it is written in many textbooks, and we can prove elementaryly this method.

10. ## Thanks Demath and Mr fantastic

Originally Posted by DeMath
Why?

The method which I offered is an analytical method, it is written in many textbooks, and we can prove elementaryly this method.

Thanks DEMATH and also Mr fantastic
cheers