The diagram shows the graph of $y = f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2005 - Paper 1

For the graph of $y = |f(x)|$, reflect any portions of the graph of $f(x)$ that are below the x-axis upwards. This means that all negative y-values become positive.

To draw the graph of $y = an(f(x))$, transform the values based on the range of $f(x)$. Identify the vertical asymptotes by finding where $f(x)$ produces values of $k \pi + \frac{\pi}{2}$.

For the graph of $y = f(|x|)$, reflect the portion of the graph for $x < 0$ across the y-axis. This way, the graph remains symmetric with respect to the y-axis.

To sketch this graph, first identify vertical asymptotes by setting the denominator to zero, giving $x^2 - 9 = 0$, thus $x = 3$ and $x = -3$. Analyze the behavior of the function as $x$ approaches these values.

Next, find the x-intercepts by setting the function equal to zero and solving. The y-intercept occurs when $x = 0$, which gives $y = 0$.

Finally, plot the key points and asymptotes, showing the general behavior of the graph in the different regions.

First, find the derivative $\frac{dy}{dx}$ using implicit differentiation. Evaluate it at the point (2, 1). The slope of the normal is the negative reciprocal of this value. Use the point-slope form of the line to find the equation of the normal.

To show this, resolve the forces in horizontal and vertical directions. Set up the equilibrium equations and substitute $mg$ and the expression for the normal force $N$. You will derive to show that the final expression indeed simplifies to $N = m \sqrt{g^2 + \frac{v^4}{r^2}}$.