Figure 1 shows the graph of $y = f(x)$, $x \in \mathbb{R}$.\nThe graph consists of two line segments that meet at the point $P$.\nThe graph cuts the $y$-axis at the point $Q$ and the $x$-axis at the points $(-3, 0)$ and $R$.\nSketch, on separate diagrams, the graphs of\n(a) $y = |f(cx)|$\n(b) $y = f(-x)$\nGiven that $f(x) = 2 - |x + 1|$,\n(c) find the coordinates of the points $P$, $Q$ and $R$.\n(d) solve $f(x) = \frac{1}{2} x$. - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 5

The graph of $y = f(-x)$ will reflect $f(x)$ across the $y$-axis. Identify the new coordinates of the peaks and intercepts based on the original graph, ensuring all aspects like symmetry are considered.

To find the coordinates of the points:\n- Point $P$: analyze the linear equations given for $f(x)$. The intersection will be $(-1, 2)$.\n- Point $Q$: substitute $x = 0$ into $f(x)$, yielding $Q(0, 1)$.\n- Point $R$: it will be located at $(1, 0)$ based on the $x$-intercept.

Set $2 - |x + 1| = \frac{1}{2} x$:\n1. For $x \geq -1$: derive $2 - (x + 1) = \frac{1}{2} x$, leading to $x = -6$. \n2. For $x < -1$: derive $2 - (-x - 1) = \frac{1}{2} x$, yielding $x = 3$.\nCombine solutions: $x = -6$ is the only valid solution in this case.