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

To sketch the graph of $y = 2f(-x) + 3$, first, we need to reflect the graph of $f(x)$ across the y-axis to get $f(-x)$. Then, we stretch this reflected graph vertically by a factor of 2 and translate it upwards by 3 units. Thus the point $P(0,11)$ would move to $P'(0, 25)$ and $Q(6,-1)$ moves to $Q'(-6, 5)$.

From the expression given, $f(x) = a|x - b| - 1$, we can deduce the following values:

• The graph crosses the y-axis at $P(0, 11)$, setting $x=0$ gives $11 = a|0 - b| - 1$. Thus $a|b| = 12$.
• The point $Q(6, -1)$ gives $-1 = a|6 - b| - 1$, leading to $a|6 - b| = 0$. Therefore, either $a = 0$ (not applicable) or $b = 6$. Since $b = 6$, we have:
• Therefore, $a = 2$.