Sorry but I'm a bit stuck at this problem with derivatives and the tangent line, I'll be stating the problem then the steps I've done to try and solve it. The problem is stated like this:
Find all points on the graph of $y = (x-2)^2$ at which the tangent line is perpendicular to the line with equation $2x - y + 2 = 0$
So I have the equation for the normal line: $y = 2(x + 1)$, so I think the equation for the tangent line is $y = -\frac{1}{2}(x + 1)$.
Since I need to get points of intersections with $(x-2)^2$, I need to get an equation for the tangent line based on it so I solve for derivatives.
$$f(x) = (x - 2)^2$$ $$f'(x) = \frac{d}{dx} (x-2)^2$$ $$f'(x) = 2(x - 2)$$
So now given these, I have another equation for the tangent line at a given point of tangency $t$:
$$y - f(t) = f'(t)(x - t)$$ $$y - (t - 2)^2 = 2(t - 2)(x - t)$$ $$y = 2(t - 2)(x - t) + (t - 2)^2$$ $$y = (t - 2)[2(x - t) + (t - 2)]$$ $$y = (t - 2)(2x - 2t + t - 2)$$ $$y = (t - 2)(2x -t - 2)$$ $$y = (t - 2)2(x - \frac{t}{2} - 1)$$ $$y = 2(t - 2)(x - \frac{t + 2}{2})$$
Now going back to the equation for the normal line, I now have two equations for the tangent line:
$$y = -\frac{1}{2}(x + 1)$$ $$y = 2(t - 2)(x - \frac{t + 2}{2})$$
Now this is where I'm stuck. The two equations seem to imply that $2(t - 2) = -\frac{1}{2}$ and $\frac{t + 2}{2} = -1$, however solving for $t$ on the first equation gives me $\frac{7}{4}$ which seem to match what's in the answer key, but the other equation gives me $t = -4$. Why is that?
Was I wrong in any of my assumptions? I know I must be missing something huge but I can't pinpoint it right now. Any help would be greatly appreciated.
Thanks.