# How do you find the asymptotes for #(9x^2 – 36) /( x^2 - 9)#?

##### 2 Answers

#### Explanation:

#"let "f(x)=(9x^2-36)/(x^2-9)# The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

#"solve "x^2-9=0rArr(x-3)(x+3)=0#

#x=+-3" are the asymptotes"#

#"Horizontal asymptotes occur as "#

#lim_(xto+-oo),f(x)toc" ( a constant)"#

#"divide terms on numerator/denominator by the highest"#

#"power of "x" that is "x^2#

#f(x)=((9x^2)/x^2-(36)/x^2)/(x^2/x^2-9/x^2)=(9-(36)/x^2)/(1-9/x^2#

#"as "xto+-oo,f(x)to(9-0)/(1-0)#

#y=9" is the asymptote"#

graph{(9x^2-36)/(x^2-9) [-40, 40, -20, 20]}

**vertical asymptotes: #x=\pm 3#**

**horizontal asymptote: #y=9#**

#### Explanation:

The given function:

**Vertical asymptotes:**

The above function will have vertical asymptote where denominator becomes zero i.e.

**Horizontal asymptotes:**

The above function will have horizontal where

Hence, **vertical asymptotes: #x=\pm 3#**

**horizontal asymptote: #y=9#**