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!

 A: 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 $(-1,-2)$
A: Suppose the $x$-coordinate of the point in question is $t$, so the $y$-coordinate is $f(t)=t^2-3$ and the point itself is $(t,t^2-3)$. You can use the fact that the slope of the tangent line there is $f'(t) = 2t$.
Since the line passes through $(t,t^2-3)$ and $(-2,0)$, compute the slope and set it equal to $f'(t)$, then find the $t$ that satisfies the equation:
$$\frac{(t^2 - 3) - 0}{(t) - (-2)} = 2t \Leftrightarrow \frac{t^2-3}{t+2}=2t \Leftrightarrow t^2-3 = 2t^2 + 4t$$
as long as $t\neq -2$. This can be rearranged as $t^2+4t+3=0$, which has solutions $t=-1$ and $t=-3$ (so the caveat $t\neq -2$ can be ignored). Geometrically, you must have $t>-2$, so the point in question is $(t,t^2-3) = \boxed{(-1,-2)}$.
A: it looks like you've already solved it! You've worked out x=-1 and put this back into the equation for y=-2x-4 to get the point P=(-1,-2) 
