Figure 1 shows part of the graph with equation $y = f(x)$, $x \in \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 5

To sketch the graph of $y = 2f(-x) + 3$, start by reflecting the original graph of $f(x)$ over the y-axis, changing $x$ to $-x$. Then, stretch the vertical distances of the graph by a factor of 2. Finally, shift the entire graph up by 3 units. Make sure the new coordinates of points $P$ and $Q$ after this transformation are shown clearly on the graph. $P$ will be located at $(0, 17)$, and $Q$ at $(-6, 5)$.

Given the information that $f(x) = a|x - b| - 1$, we know that the minimum value of $f(x)$ at $Q(6, -1)$ gives us:
$-1 = a|6 - b| - 1$. This simplifies to $a|6 - b| = 0$, implying that $|6 - b| = 0$. Thus, $b = 6$.

From the value at $P(0, 11)$, we can evaluate:
$11 = a|0 - 6| - 1$, which simplifies to $11 = 6a - 1$, leading to $6a = 12$, thus $a = 2$.

Therefore, $a = 2$ and $b = 6$.