How do you find the horizontal and vertical asymptotes to the curve of y =(e^2x +1)/(e^x -2) ?

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Answer 1 · 2017-07-14T10:20:59

The vertical asymptote is #y=ln2#
The horizontal asymptote is #y=-1/2#

The equation of the curve is

#y=(e^(2x)+1)/(e^x-2)#

The denominator is #=0#, when

#e^x-2=0#

#e^x=2#

#x=ln2#

To calculate the vertical asymptotes , we calculate

#lim_(x->ln2,x < ln2)y=lim_(x->ln2,x < ln2)(e^(2x)+1)/(e^x-2)=5/0^(- )=-oo#

As #e^(2*ln2)+1=4+1=5# and #e^(ln2)-2=2^(- )-2=0^-#

#lim_(x->ln2,x > ln2)y=lim_(x->ln2,x > ln2)(e^(2x)+1)/(e^x-2)=5/0^(+ )=+oo#

As #e^(2*ln2)+1=4+1=5# and #e^(ln2)-2=2^(+ )-2=0^+#

Therefore,

the vertical asymptote is #y=ln2#

To calculate the horizontal asymptotes , we calculate

#lim_(x->+oo)y=lim_(x->+oo)(e^(2x)+1)/(e^x-2)=lim_(x->+oo)e^(2x)/e^x=lim_(x->+oo)e^x=+oo#

#lim_(x->-oo)y=lim_(x->-oo)(e^(2x)+1)/(e^x-2)=lim_(x->-oo)(0^+ +1)/(0^+ -2)=-1/2#

As #lim_(x->-oo)e^(2x)=0^+# and #lim_(x->-oo)e^(x)=0^+#

Therefore,

the horizontal asymptote is #y=-1/2#

graph{(e^(2x)+1)/(e^x-2) [-38.55, 34.5, -19.86, 16.7]}