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.
Given curve $y=kx^2+(3k-1)x- 1$ and curve $y=(k+1)x-11$ are intersect each
other exactly at one ponit (Means only one value of $x$ and Corrosponding value of $y$)
which is also called Condition of tangency.
So Equating $y\;,$ We get $$kx^2+(3k-1)x-1 = (k+1)x-11$$
So $$\displaystyle kx^2+2(k-1)x+10=0\Rightarrow x=\frac{-2(k-1)x\pm\sqrt{4(k-1)^2-4\cdot k\cdot 10}}{2k}$$
Now for only one value of $x\;,$ We have $$\displaystyle 4(k-1)^2-4\cdot k\cdot 10=0$$
So $$\displaystyle(k-1)^2-10k=0\Rightarrow k^2-12k+1=0\Rightarrow k = \frac{12\pm\sqrt{140}}{2} = 6\pm \sqrt{35}$$
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.
