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