Let $P(x) = x^3 + 3x^2 - 13x + 6.$ (i) Show that $P(2) = 0.$ (ii) Hence, factor the polynomial $P(x)$ as $A(x)B(x)$, where $B(x)$ is a quadratic polynomial - HSC - SSCE Mathematics Extension 1 - Question 11 - 2020 - Paper 1

Since $P(2) = 0$, we know that $(x - 2)$ is a factor of $P(x)$. To find the other factor, we can perform polynomial long division:

Dividing $P(x)$ by $(x - 2)$:

1. Write $P(x) = x^3 + 3x^2 - 13x + 6$.
2. Divide the leading term $x^3$ by $x$ to get $x^2$.
3. Multiply $(x - 2)$ by $x^2$: $x^2(x - 2) = x^3 - 2x^2$
4. Subtract this from $P(x)$: $P(x) - (x^3 - 2x^2) = 5x^2 - 13x + 6$
5. Divide the leading term $5x^2$ by $x$ to get $5x$.
6. Multiply $(x - 2)$ by $5x$: $5x(x - 2) = 5x^2 - 10x$
7. Subtract again: $5x^2 - 13x + 6 - (5x^2 - 10x) = -3x + 6$
8. Divide $-3x$ by $x$ to get $-3$.
9. Multiply $(x - 2)$ by $-3$: $-3(x - 2) = -3x + 6$
10. Subtract one last time: $-3x + 6 - (-3x + 6) = 0.$

Thus, we can express $P(x)$ as: $P(x) = (x - 2)(x^2 + 5x - 3).$

Here, $A(x) = (x - 2)$ and $B(x) = (x^2 + 5x - 3)$.

Two vectors are perpendicular if their dot product is zero. The dot product of the vectors is calculated as follows:

$\begin{pmatrix} a \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 2a - 3 \\ 2 \end{pmatrix} = a(2a - 3) + (-1)(2) = 2a^2 - 3a - 2 = 0.$

To find the values of $a$, we can solve the quadratic equation:

$2a^2 - 3a - 2 = 0.$

Using the quadratic formula: $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-2)}}{2(2)} = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4}.$

This results in: $a = \frac{8}{4} = 2$ and $a = \frac{-2}{4} = -\frac{1}{2}.$

Therefore, the values of $a$ that make the vectors perpendicular are $a = 2$ and $a = -\frac{1}{2}$.

To sketch the graph of $y = \frac{1}{f(x)}$, we first analyze the original graph of $y = f(x)$. The polynomial $f(x) = x^3 + 3x^2 - 13x + 6$ has one local maximum and one local minimum, which will impact the graph of the reciprocal function.

1. Identify the zeroes of $f(x)$ which correspond to vertical asymptotes in $y = \frac{1}{f(x)}$.
2. For any $x$ where $f(x)$ is positive, $y = \frac{1}{f(x)}$ will be a decreasing function, and for any $x$ where $f(x)$ is negative, it will be increasing.
3. The graph will approach infinity as $f(x)$ approaches 0 and will tend to zero as $f(x)$ increases to positive or negative infinity.
4. Mark the approximate locations of asymptotes and critical points from the graph of $f(x)$. Include key features such as intersections and the axis behavior.