How do you find the asymptotes for #y = (7x-5)/(2-5x)#?

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Answer 1 · 2016-02-22T07:43:42

The asymptotes are #x=2/5# vertical asymptote
#y=-7/5# horizontal asymptote

Take the limit of y as x approaches #oo#

#lim_(x->oo) y=lim_(x->oo) (7x-5)/(-5x+2)=lim_(x->oo) (7-5/x)/(-5+2/x)=-7/5#

#x=-7/5#

Also if you solve for x in terms of y,

#y=(7x-5)/(-5x+2)#

#y(-5x+2)=7x-5#

#-5xy+2y=7x-5#

#2y+5=7x+5xy#

#2y+5=x(7+5y)#

#x=(2y+5)/(5y+7)#

take now the limit of x as y approaches #oo#

#lim_(y->oo) x=lim_(y->oo) (2y+5)/(5y+7)=lim_(y->oo) (2+5/y)/(5+7/y)=2/5#

#y=2/5#

kindly see the graph.

graph{y=(7x-5)/(-5x+2)[-20,20,-10,10]}

have a nice day!