(a) The diagram shows the graph of $y = f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2006 - Paper 1

For the graph of $y = \frac{1}{f(x)}$, we consider the behavior of the graph based on the asymptotes of $f(x)$. Since $f(x)$ has a horizontal asymptote at $y = 2$, $y = \frac{1}{f(x)}$ will have a horizontal asymptote at $y = \frac{1}{2}$. Additionally, where $f(x)$ approaches 0, the graph of $y = \frac{1}{f(x)}$ will tend towards ±∞. Points where $f(x)$ has local maxima or minima need to be marked carefully.

To sketch the graph of $y = x f(x)$, we must consider the behavior of both $x$ and $f(x)$. This graph will cross the origin since $f(0) = 0$, and it will also exhibit symmetry around the origin. As $x$ approaches ±∞, the function will amplify the behavior of $f(x)$, resulting in a curve that stretches outwards while being influenced by the horizontal asymptote at $y = 2$.

To find the intersection points of the hyperbola with the $x$-axis, set $y = 0$ in the hyperbola equation: [ \frac{x^2}{144} - \frac{0^2}{25} = 1 ]
This simplifies to [ \frac{x^2}{144} = 1 ] leading to [ x^2 = 144 ] and therefore the intersection points are $(12, 0)$ and $(-12, 0)$.

For the hyperbola given by [ \frac{x^2}{144} - \frac{y^2}{25} = 1, ] we identify $a^2 = 144$ and $b^2 = 25$. The distance to the foci is given by $c = \sqrt{a^2 + b^2} = \sqrt{144 + 25} = \sqrt{169} = 13.$ Thus, the coordinates of the foci are $(13, 0)$ and $(-13, 0)$.

The equations for the directrices of the hyperbola can be found using the formula $x = \pm \frac{a^2}{c}$, yielding: [ x = \pm \frac{144}{13} ] The asymptotes are given by the equations: [ y = \pm \frac{b}{a} x ] leading to: [ y = \pm \frac{5}{12} x. ]

To find the values of $a$ and $b$, we note that since $a + ib$ and $a + 2ib$ are zeros of the polynomial $P(x)$, their complex conjugates must also be zeros. Equating coefficients with the expanded form of the factors leads to a system of equations which can be solved to find the specific values of $a$ and $b$.

By substituting the found values for $a$ and $b$ into the factors corresponding to the complex roots, we can express $P(x)$ in its factored form. The resulting quadratics can be of the form $(x - (a + ib))(x - (a - ib))(x - (a + 2ib))(x - (a - 2ib))$ simplifying to two quadratic factors with real coefficients. The explicit form needs to be computed to provide the final answer.