a)
\(\[\begin{align} & f(x)=-{{x}^{2}}+4x-3 \\ & f(2)=-{{2}^{2}}+4\cdot 2-3 \\ & f\left( 2 \right)=-4+8-3 \\ & f\left( 2 \right)=1 \\ \end{align}\] \)

Portanto, prova-se que (2,1) pertence à parábola.

b)
\(\[\begin{align} & {{T}_{2}}\left( x,y \right)=\left( x+y,-2x+4y \right) \\ & {{T}_{2}}(x,y)=\lambda \cdot (x,y) \\ & (x,y)\ne (0,0): \\ & (x+y,-2x+4y)=(\lambda \cdot x,\lambda \cdot y) \\ & \left\{ \begin{array}{*{35}{l}} x+y=\lambda \cdot x \\ -2x+4y=\lambda \cdot y \\ \end{array} \right. \\ & \left\{ \begin{array}{*{35}{l}} (1-\lambda )x+y=0 \\ -2x+(4-\lambda )y=0 \\ \end{array} \right. \\ & \text{det}\left[ \begin{matrix} (1-\lambda ) & 1 \\ -2 & (4-\lambda ) \\ \end{matrix} \right]=0 \\ & (1-\lambda )(4-\lambda )-1\cdot (-2)=0(1-\lambda )(4-\lambda )+2=0{{\lambda }^{2}}-5\lambda +6=0 \\ & \text{As soluc }\!\!\tilde{\mathrm{o}}\!\!\text{ es para a equac }\!\!\tilde{\mathrm{a}}\!\!\text{ o s }\!\!\tilde{\mathrm{a}}\!\!\text{ o:} \\ & \lambda =\frac{1}{2}\left( 5\pm \sqrt{{{(-5)}^{2}}-4\cdot 1\cdot 6} \right) \\ & \lambda =\frac{1}{2}\left( 5\pm 1 \right) \\ & {{\lambda }_{1}}=\frac{1}{2}(5+1)=3{{\lambda }_{2}}=\frac{1}{2}(5-1)=2 \\ & \\ & \\ \end{align}\] \)

c)

\(\[\begin{align} & T\_2\left( x,y \right)=\left( x+y,-2x+4y \right) \\ & voltamos\text{ }\grave{a}\text{ }equa\tilde{a}o\text{ }de\text{ }autovetores: \\ & {{T}_{2}}(x,y)=\lambda \cdot (x,y) \\ & para\text{ }o\text{ }autovalor\text{ }\lambda \text{ }=\text{ }3: \\ & {{T}_{2}}(x,y)=\lambda \cdot (x,y)(x+y,-2x+4y) \\ & 3\cdot (x,y) \\ & \left\{ \begin{array}{*{35}{l}} x+y=3x \\ -2x+4y=3y \\ \end{array} \right.e\left\{ \begin{array}{*{35}{l}} y=2x \\ y=2x \\ \end{array} \right. \\ & {{v}_{1}}=(1,2),\text{para}\lambda =3 \\ & {{T}_{2}}(x,y)=\lambda \cdot (x,y)(x+y,-2x+4y)=2\cdot (x,y) \\ & \left\{ \begin{array}{*{35}{l}} x+y=2x \\ -2x+4y=2y \\ \end{array} \right. \\ & \left\{ \begin{array}{*{35}{l}} y=x \\ y=x \\ \end{array} \right. \\ & {{v}_{2}}=(1,1),\text{para}\lambda =2. \\ & \\ \end{align}\] \)