Đáp án:
a) tại x=2
$\begin{array}{l}
\Rightarrow y = - {x^2} = - {2^2} = - 4\\
\Rightarrow M\left( {2; - 4} \right)
\end{array}$
Gọi phương trình d có dạng: y=a.x+b
$\begin{array}{l}
\Rightarrow - {x^2} = a.x + b\\
\Rightarrow {x^2} + a.x + b = 0\\
\Rightarrow \left\{ \begin{array}{l}
\Delta = 0\\
- 4 = 2.a + b
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{a^2} - 4.b = 0\\
b = - 4 - 2a
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{a^2} - 4.\left( { - 4 - 2a} \right) = 0\\
b = - 4 - 2a
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{a^2} + 8a + 16 = 0\\
b = - 4 - 2a
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = - 4\\
b = - 4 - 2a
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = - 4\\
b = - 4 - 2.\left( { - 4} \right) = 4
\end{array} \right.\\
Vay\,\left( d \right):y = - 4x + 4\\
b)\left( {{d_2}} \right)//\left( d \right):y = 2x + 3\\
\Rightarrow {d_2}:y = 2x + b\left( {b \ne 3} \right)\\
\Rightarrow - {x^2} = 2x + b\\
\Rightarrow {x^2} + 2x + b = 0\\
\Rightarrow \Delta ' = 0\\
\Rightarrow 1 - b = 0\\
\Rightarrow b = 1\left( {tm} \right)\\
\Rightarrow \left( {{d_2}} \right):y = 2x + 1\\
c)\left( {{d_a}} \right):y = a.x + b\\
A\left( { - 1;3} \right) \in {d_a}\\
\Rightarrow 3 = - a + b\\
\Rightarrow b = a + 3\\
- {x^2} = a.x + b\\
\Rightarrow {x^2} + a.x + b = 0\\
\Rightarrow \Delta = 0\\
\Rightarrow {a^2} - 4b = 0\\
\Rightarrow {a^2} - 4.\left( {a + 3} \right) = 0\\
\Rightarrow {a^2} - 4a - 12 = 0\\
\Rightarrow \left( {a - 6} \right)\left( {a + 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
a = 6 \Rightarrow b = 9\\
a = - 2 \Rightarrow b = 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{d_a}:y = 6x + 9\\
{d_a}:y = - 2x + 1
\end{array} \right.
\end{array}$
