How do you find vertical, horizontal and oblique asymptotes for #4/(1 -x^2)#?
1 Answer
vertical asymptotes at x = ± 1
horizontal asymptote at y = 0
Explanation:
The denominator of the function cannot be zero as this would make the function 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 :
#1-x^2=0rArr(1-x)(1+x)=0#
#rArrx=-1" and " x=1" 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)=(4/x^2)/(1/x^2-x^2/x^2)=(4/x^2)/(1/x^2-1)# as
#xto+-oo,f(x)to0/(0-1)#
#rArry=0" is the asymptote"# Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 0, denominator-degree 2 ) Hence there are no oblique asymptotes.
graph{(4)/(1-x^2) [-10, 10, -5, 5]}
