How do you find the vertical, horizontal and slant asymptotes of: #y=(2x )/ (x-5)#?
1 Answer
vertical asymptote
horizontal asymptote
Explanation:
For this rational function (fraction) the denominator cannot be zero. This would lead to division by zero which is undefined. By setting the denominator equal to zero and solving for
#x# we can find the value that#x# cannot be. If the numerator is also non-zero for such a value of#x# then this must be a vertical asymptote.solve :
#x - 5 = 0 rArr x = 5# is the asymptoteHorizontal asymptotes occur as
#lim_(xto+-oo) ytoc" (a constant)"# divide terms on numerator/denominator by
#x#
#((2x)/x)/(x/x-5/x)=2/(1-5/x)# as
#xto+-oo, yto2/(1-0)#
#rArry=2" is the asymptote"# Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both degree 1 ) Hence there are no slant asymptotes.
graph{(2x)/(x-5) [-20, 20, -10, 10]}
