$T$ is the triangle formed by (0,0),(1,0) and (0,2) in the $uv$ plane. The transformation is defined by $x= 2u+v ,y=u^2−v$.
I'm completely lost. I've been trying to write $u$ and $v$ as a function of $x$ and $y$ but I can't do it.
I've been able to figure out that the triangle in the $uv$ plane is defined by $u>0, 2 \ge v + 2u$ and thus $0 \le x \le 2$. However I can't complete the exercise. The answer is what I've got and $-x \le y \le (\frac x 2)^2$
Let the $\triangle AOB$ be such that $A(1,0), B(0,2)$ and $O(0,0)$.
Parametric representation of the line segment $OA$ is given by $u=t$ and $v=0$ with $0 \leq t \leq 1$. So the image of this segment under the given transformation is given by: \begin{align*} x & = 2t+0=2t\\ y&=t^2-0=t^2. \end{align*} This means the line segment $OA$ is mapped to the curve $y=\frac{x^2}{4}$, where $0 \leq x \leq 2$.
Parametric representation of the line segment $OB$ is given by $u=0$ and $v=s$ with $0 \leq s \leq 2$. So the image of this segment under the given transformation is given by: \begin{align*} x & = s\\ y&=-s. \end{align*} This means the line segment $OA$ is mapped to the line $y=-x$, where $0 \leq x \leq 2$.
Parametric representation of the line segment $BA$ is given by $u=c$ and $v=2-2c$ with $0 \leq c \leq 1$. So the image of this segment under the given transformation is given by: \begin{align*} x & =2\\ y&=c^2+2c-2. \end{align*} This means the line segment $OA$ is mapped to the vertical line $x=2$, where $-2 \leq y \leq 1$.
Thus the $\triangle AOB$ will be mapped to the region $A'OB'$, which is given by $-x \leq y \leq \frac{x^2}{4}$, where $0 \leq x\leq 2$.
Here is a image that will help you.