Need help with tangents to a quadratic The quadratic $y=kx^2+(3k-1)x-1$ and the straight line $y=(k+1)x-11$ meet. 
Find the range of value(s) of $k$ such that the line is a tangent to the curve.
Got this question for school. Seems really simple and it's a non-calculator question but I'm not sure how to go about it. 

HINT:
Equate the values of $y$ to form a Quadratic Equation in $x$, each of the two roots represents the abscissa of the respective intersection.
The straight line $y=(k+1)x−11 $ says that each abscissa has a unique ordinate associated
For tangency, the two values of $x$ must coincide.
