There exists exactly one positive value of k such that the line y = kx intersects the parabola
y = x2 + x + 4 at exactly one point. Compute the intersection point.	Answers
A) (2,10)  B) (5,10)  C) (7,10)  D) (3,10)  E) (1,10)
If the line is tangent, then we must have that x2 + x + 4 = kx has a unique solution   in x, or x2 + (1 - k)x + 4 = 0 has a unique solution. Thus, (1 - k)2- 16 = 0, or (1- k) = 4.   Therefore, k = -3 or k = 5.   Taking k = 5, we have that y = 5x intersects y = x2 + x + 4 when x = 2, and the intersection   point is (2, 10) .   correct answer (2,10)
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There exists exactly one positive value of k such that the line y = kx intersects the parabola
y = x2 + x + 4 at exactly one point. Compute the intersection point.