# 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?

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

$D = \sqrt{{b}^{2} - 4 a c} = \sqrt{{\left(- 1\right)}^{2} - 4 k} = \sqrt{1 - 4 k}$

but we want $D = 0 \implies 1 - 4 k = 0 \implies k = \frac{1}{4}$

Now for $k = \frac{1}{4}$ we have that

$y = \frac{2 x + 1}{{x}^{2} - x + \frac{1}{4}} = \frac{2 x + 1}{x - \frac{1}{2}} ^ 2$

Now we see that for $x \to \frac{1}{2}$ $y \to \infty$ hence $x = \frac{1}{2}$ vertical asymptote.

There is another possibility that results in a single vertical asymptote: If x^2−x+k is divisible by $2 x + 1$ 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