In geometry, asymptote is a line in such a way that the distance between the line and the curve tends to zero when they approaches to infinity. Mainly there are two types of asymptotes; *horizontal and vertical asymptotes*. The formula for horizontal and vertical asymptotes are,

- Write down the given equation in y= form
- Find out the ratio between the highest exponents of x in numerator and denominator

Note:

If the exponent of denominator is larger than that of numerator, horizontal asymptote, y=0. If the exponent of numerator is high, there is no horizontal asymptote.

**Vertical Asymptote:**

- Set the denominator equal to zero and resolve the equation.

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Solved problems of asymptotes are given below:

### Solved Examples

**Question 1: **Find out the horizontal and vertical asymptotes of $\frac{2^{3}+5x+3}{3x^{3}-9}$ ?

** Solution: **

Given function is,

y = $\frac{2^{3}+5x+3}{3x^{3}-9}$

Highest exponent of numerator and denominator is 3. The coefficients are 2 and 3 respectively.

Horizontal asymptote is, $\frac{2}{3}$

Vertical asymptote can be calculated as,

3$x^{3}$ - 9 = 0

3$x^{3}$ = 9

$x^{3}$ = 3

x = $\sqrt[3]{3}$

x = 1.4422495

**Question 2: **Calculate the horizontal and vertical asymptotes of $\frac{x+2}{x^{2}-4}$ ?

** Solution: **

The given function is,

y = $\frac{x+2}{x^{2}-4}$

The exponent of denominator is higher than that of numerator. So horizontal asymptote is y = 0.

Vertical asymptote,

$x^{2}$ - 4 = 0

$x^{2}$ = 4

x = $\sqrt{4}$

x = 2

Given function is,

y = $\frac{2^{3}+5x+3}{3x^{3}-9}$

Highest exponent of numerator and denominator is 3. The coefficients are 2 and 3 respectively.

Horizontal asymptote is, $\frac{2}{3}$

Vertical asymptote can be calculated as,

3$x^{3}$ - 9 = 0

3$x^{3}$ = 9

$x^{3}$ = 3

x = $\sqrt[3]{3}$

x = 1.4422495

The given function is,

y = $\frac{x+2}{x^{2}-4}$

The exponent of denominator is higher than that of numerator. So horizontal asymptote is y = 0.

Vertical asymptote,

$x^{2}$ - 4 = 0

$x^{2}$ = 4

x = $\sqrt{4}$

x = 2

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