Question 3 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 3 - 2006 - Paper 1

For the graph of $y = \frac{1}{f(x)}$, we recognize that where $f(x)$ crosses the x-axis, $y = \frac{1}{f(x)}$ will have vertical asymptotes. Consequently, if $f(x)$ is positive, then the reciprocal will also be positive and vice versa.

Since $f(x)$ approaches $2$ as a horizontal asymptote, $\frac{1}{f(x)}$ approaches $\frac{1}{2}$ as $x$ tends to both directions. The key behavior will show that sections of the graph are reflected due to the reciprocal nature.

To sketch the graph of $y = x f(x)$, we recognize that this graph will be dependent on both $x$ and $f(x)$.

For $x > 0$, the graph behaves similarly to $f(x)$, scaled by $x$. For $x < 0$, the graph will reflect across the origin where $f(x)$ is negative. Here, the graph can have zero crossings at the zeros of $f(x)$ but will adapt into broader ranges as $x$ increases or decreases. This sketch will show a characteristic 'flattening' as the horizontal asymptote couples with the behavior of $x$ in both directions.

To find the points where the hyperbola intersects the $x$-axis, we set $y = 0$ in the hyperbola equation:

$\frac{x^2}{144} - \frac{0^2}{25} = 1$ This reduces to $\frac{x^2}{144} = 1$ Leading to $x^2 = 144 o x = \pm 12.$
Thus, the intersections occur at the points $(12, 0)$ and $(-12, 0)$.

The coordinates of the foci for a hyperbola are found using the formula $c = \sqrt{a^2 + b^2}$, where $a = 12$ and $b = 5$. Evaluating this gives:

$c = \sqrt{144 + 25} = \sqrt{169} = 13.$ Thus, the foci are located at: $(13, 0)$ and $( -13, 0).$

The equations of the directrices for the hyperbola are given by $x = \pm \frac{a^2}{c} = \pm \frac{144}{13}$. The directrices therefore are:

$x = \frac{144}{13}$ and $x = -\frac{144}{13}.$

For the asymptotes of the hyperbola: $y = \pm \frac{b}{a} (x) = \pm \frac{5}{12} x.$

To find the values of $a$ and $b$, we equate the polynomial to have appropriate roots for the complex zeros. Identifying symmetry in the coefficients helps to calculate directly while substituting complex conjoinments leads to equations for $a$ and $b$. Solving these yields:

$a = 3 ext{ and } b = 1.$

Given the identified values of $a$ and $b$, we can express $P(x)$ as:

$P(x) = (x^2 - 6x + 10)(x^2 - 6x + 13).$

These quadratics yield the required factorization with real coefficients.