Figure 1 shows the graph of $y = f(x)$, $-5 \leq x \leq 5$ - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 5

For this graph, we take the absolute value of the function $f(x)$. Any segments of the graph that fall below the x-axis will be reflected above the x-axis. The maximum turning point M(2, 4) remains the maximum point. Find points where $f(x)$ is negative and replace their $y$ values with their absolute values. Ensure any local minima below the x-axis are accurately reflected above.

The function $g(x) = e^{x^2}$ is even, meaning $g(-x) = g(x)$. Therefore, we only need to graph for $x \geq 0$ and then reflect that section across the y-axis. For $x = 0$, $g(0) = e^0 = 1$. As $x$ increases, $g(x)$ also increases rapidly due to the $e^{x^2}$ term. The turning point at $x = 2$ in the original function translates into the new graph symmetrically.

In the graph of $y = f(x)+3$, the maximum turning point is M'(2, 7). For $y = |f(x)|$, the maximum turning point remains at M(2, 4) as it is above the x-axis. In the case of $y = g(|x|)$, since the graph is symmetric about the y-axis and the turning point is evaluated at $x = 2$, the points M(2, $e^{4}$) and M(-2, $e^{4}$) can be noted where the graph changes direction.