Question

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

Answer 1

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]}