How can I find the coordinates of a point given a function, graph, and Tangent line point? So I thought I was doing it right but I can't be. I keep getting a positive slope for the TL and the coords I get don't make sense for the graph of the function. 
I took a picture of my work. If someone could please help, I'd greatly appreciate it. Thanks!


Let the required point on the curve be $(k,k^2-3)$. Then the slope of the tangent at that point equals $2k$. Further we know it passes through two points $(-2,0)$ and $(k,k^2-3)$. So, lets define it by $L:y = 2kx+c$. 
Substituting the $2$ points in the equation, we get $c = 4k$ and $k^2+4k+3=0$ which gives $k=-3$ or $-1$. So the required points on the graph are $(-3,6)$ and $(-1,-2)$
