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

2 Answers
Jul 28, 2018

Answer:

#"vertical asymptotes at "x=+-3#
#"horizontal asymptote at "y=9#

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]}

Answer:

vertical asymptotes: #x=\pm 3#

horizontal asymptote: #y=9#

Explanation:

The given function:

#f(x)=\frac{9x^2-36}{x^2-9}#

Vertical asymptotes:

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

#\therefore x^2-9=0#

#x^2=9#

#x=\pm 3#

Horizontal asymptotes:

The above function will have horizontal where

#y=\lim_{x\to \pm \infty}\frac{9x^2-36}{x^2-9}#

#y=\lim_{x\to \pm \infty}\frac{9-36/x^2}{1-9/x^2}#

#y=\frac{9-0}{1-0}#

#y=9#

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

horizontal asymptote: #y=9#