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

1 Answer
Jul 14, 2017

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

Explanation:

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]}