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:
horizontal asymptote:
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:
horizontal asymptote: