(12 points) Consider the curve $C: x=t^{3}+6 t+1, y=2 t-t^{2}$ (a) (4 points) Find the slope of the line that is tangent to the curve at the point given by $t=-1$ (b) (4 points) Find all points $(x, y)$ on $\mathrm{C}$ where the tangent is horizontal or vertical.

Step 1 ：Let's denote the parameter as \(t\), the x-coordinate as \(x = t^3 + 6t + 1\), and the y-coordinate as \(y = 2t - t^2\).

Step 2 ：We can find the derivatives of x and y with respect to \(t\) as \(dx/dt = 3t^2 + 6\) and \(dy/dt = 2 - 2t\) respectively.

Step 3 ：The derivative of y with respect to x, \(dy/dx\), which gives the slope of the tangent line at any point on the curve, can be found as \(dy/dx = (dy/dt) / (dx/dt) = (2 - 2t) / (3t^2 + 6)\).

Step 4 ：Substituting \(t = -1\) into \(dy/dx\), we find the slope of the tangent line at the point given by \(t = -1\) to be \(4/9\).

Step 5 ：A horizontal tangent line occurs when \(dy/dx = 0\), which gives \(t = 1\). Substituting \(t = 1\) into the equations for x and y, we find the point on the curve where the tangent is horizontal to be \((8, 1)\).

Step 6 ：A vertical tangent line occurs when \(dx/dt = 0\), but this equation has no real solutions, so there are no points on the curve where the tangent is vertical.

Step 7 ：Final Answer: (a) The slope of the line that is tangent to the curve at the point given by \(t = -1\) is \(\boxed{\frac{4}{9}}\). (b) The point on the curve \(C\) where the tangent is horizontal is \(\boxed{(8, 1)}\). There are no points on the curve where the tangent is vertical.