Use the Question 11 Writing Booklet (a) Let $P(x) = x^3 + 3x^2 - 13x + 6$ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2020 - Paper 1

Given that $P(2) = 0$, we can factor $P(x)$ as follows:

Using synthetic division to divide $P(x)$ by $(x - 2)$:

egin{array}{r|rrrr} 2 & 1 & 3 & -13 & 6 \\ & & 2 & 10 & -6 \\ ext{------------------} \\ & 1 & 5 & -3 & 0 \\ ext{Thus, } \ P(x) = (x - 2)(x^2 + 5x - 3).

So, we have factored the polynomial as $A(x) = (x - 2)$ and $B(x) = (x^2 + 5x - 3)$.

Vectors are perpendicular if their dot product is zero:

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

Calculating the dot product:

$$a(2a - 3) + (-1)(2) = 0 \\ 2a^2 - 3a - 2 = 0.$$

Next, we can solve this quadratic using the quadratic formula:

$$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-2)}}{2(2)}.$$

This simplifies to:

$$a = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4}.$$

Thus, the solutions are:

$$a = 2 \text{ or } -\frac{1}{2}.$$

To sketch the graph of $y = \frac{1}{f(x)}$, we first need to understand the behavior of $f(x)$, which is a downward-opening parabola with a minimum point.

At the minima, the value of $f(x)$ approaches zero, leading to vertical asymptotes in $y = \frac{1}{f(x)}$. The asymptotes occur at the $x$-values where $f(x) = 0$, which we found earlier occurs at $x = 2$.

Also, as $x$ approaches the minima of $f(x)$, $y = \frac{1}{f(x)}$ approaches infinity. Between the roots and the location of the minima, the graph of $y = \frac{1}{f(x)}$ will exhibit a downward curve.

Therefore, the sketch will have the necessary features including vertical asymptotes as described.