The curve $C$ is parameterized by $x(t)=t^{2}-2 t$ and $y(t)=t^{3}-6 t^{2}$.
1. There is exactly one point $(a, b)$ at whch the tangent line to $C$ at this point is vertical. The first coordinate $a$ is
2. There are exactly two points $\left(a_{1}, b_{1}\right)$ and $\left(a_{2}, b_{2}\right)$ on this curve for which the tangent line to $C$ is horizontal. List these points so that $a_{1} \leq a_{2}$ : The first coordinate $a_{1}$ of the first point is and the first coordinate $a_{2}$ of the second point is
3. For each of the following points, we would like to know about the concavity of the curve at those points. Answer with "U" for concave up, "D" for concave down, "I" for an inflection point, and "N" for none of these.
The concavity at $(3,5)$ is and the concavity at $(15,27)$ is

Step 1 ：Given the curve $C$ is parameterized by $x(t)=t^{2}-2 t$ and $y(t)=t^{3}-6 t^{2}$.

Step 2 ：We need to find the point $(a, b)$ at which the tangent line to $C$ at this point is vertical. This occurs when the derivative of $y$ with respect to $x$, $\frac{dy}{dx}$, is undefined.

Step 3 ：To find $\frac{dy}{dx}$, we first find $\frac{dx}{dt}$ and $\frac{dy}{dt}$, and then divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$.

Step 4 ：Calculating $\frac{dx}{dt}$, we get $2t - 2$.

Step 5 ：Calculating $\frac{dy}{dt}$, we get $3t^{2} - 12t$.

Step 6 ：Dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$, we get $\frac{3t^{2} - 12t}{2t - 2}$.

Step 7 ：We set $\frac{dy}{dx}$ equal to infinity (or undefined) and solve for $t$. This gives us $t = 1$.

Step 8 ：Substituting this value of $t$ back into the equation for $x(t)$, we get $a = -1$.

Step 9 ：Final Answer: The first coordinate $a$ at which the tangent line to $C$ at this point is vertical is \(\boxed{-1}\).