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?
I would be careful with terminology. First, one can only talk about the intersection of graphs (which are geometrical figures), not equations. Second, we can draw all points satisfying two equations and , but this would not be the set of points satisfying . For example, points satisfying 2x + 3y = 0 and x + y = 0 are lines that intersect at just one point (0, 0), but points satisfying x + 2y = 0 fill a whole line. What we have in the original problem are equations and , to be sure, but not arbitrary equations. Rather, they are definitions of functions with only y in the left-hand side.
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.
I don't have a graphing calculator, so I am not sure. There are many possible reasons. E.g., you may need to adjust the window size, write "2*sqt(2)" instead of "2sqt2," etc. Try using 1.41 instead of sqt2.