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?

1 Answer

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