3
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e93e3c51-ae2b-420b-abb8-bf0c483caff8-3_1270_1219_326_463}
\captionsetup{labelformat=empty}
\caption{Fig. 12}
\end{figure}
Fig. 12 shows the graph of \(y = \frac { 1 } { x - 2 }\).
- Draw accurately the graph of \(y = 2 x + 3\) on the copy of Fig. 12 and use it to estimate the coordinates of the points of intersection of \(y = \frac { 1 } { x - 2 }\) and \(y = 2 x + 3\).
- Show algebraically that the \(x\)-coordinates of the points of intersection of \(y = \frac { 1 } { x - 2 }\) and \(y = 2 x + 3\) satisfy the equation \(2 x ^ { 2 } - x - 7 = 0\). Hence find the exact values of the \(x\)-coordinates of the points of intersection.
- Find the quadratic equation satisfied by the \(x\)-coordinates of the points of intersection of \(y = \frac { 1 } { x - 2 }\) and \(y = - x + k\). Hence find the exact values of \(k\) for which \(y = - x + k\) is a tangent to \(y = \frac { 1 } { x - 2 }\). [4]