It does look tangent.
Hint: consider the difference of the two functions and find when it has a single root.
Ok so I got x^(2) (-2-k)x + 2 = 0
Ultimately I got k^(2) +4k -4 = 0 I used quadratic formula and came up with the correct answer. So theoretically, when you subtract the two equations, the solution(s) of that equation is the intersection(s) of both equations? And with resepect to the calculator, did I plug it on wrongly? Why did they not appear to intersect?
So, the correct statement is, If you have functions and , then the solutions of the equation give the (x-coordinates of the) intersection points of the graphs of the two functions.