For what values of #a# and #m# does #f(x) = (2x^m)/(x+a)# have a horizontal asymptote at y=2 and a vertical asymptote at x=1?
1 Answer
Aug 4, 2017
Explanation:
The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.
#"if "a=-1#
#rArrx-1=0rArrx=1" is the asymptote"#
#"for a horizontal asymptote "y=2#
#"the degrees of the numerator and denominator must be "#
#"equal"#
#rArrm=1#
#rArrf(x)=(2x)/(x-1)#
#"horizontal asymptotes occur as"#
#lim_(xto+-oo),f(x)toc" (a constant)"#
#"divide terms on numerator/denominator by x"#
#f(x)=((2x)/x)/(x/x-1/x)=2/(1-1/x)# as
#xto+-oo,f(x)to2/(1-0)#
#rArry=2" is the asymptote"#
graph{(2x)/(x-1) [-10, 10, -5, 5]}