Question

There is only one vartical asymptote of the curve: `y=(2x+1)/(x^2-x+k)` What is the sum of values that `k` can have?

Answer 1

Refer to explanation

Explanation:

Since there is only one vertical asymptote that means that the trinomial `x^2-x+k` must have one double root hence its discriminant must be zero .

No the discriminant is equal to

`D=sqrt(b^2-4ac)=sqrt((-1)^2-4k)=sqrt(1-4k)`

but we want `D=0=>1-4k=0=>k=1/4`

Now for `k=1/4` we have that

`y=(2x+1)/(x^2-x+1/4)=(2x+1)/(x-1/2)^2`

Now we see that for `x->1/2` `y->oo` hence `x=1/2` vertical asymptote.

There is another possibility that results in a single vertical asymptote: If `x^2−x+k` is divisible by `2x+1` then there will be just the one asymptote. This occurs when `k=−3/4` and `x^2−x−3/4=(2x+1)(1/2x−3/4)` (*)

(*) Credits to George C. for this addition