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

Since $P(2) = 0$, $(x - 2)$ is a factor of $P(x)$.

Now, we will perform polynomial long division of $P(x)$ by $(x - 2)$:

1. Divide the leading term: (\frac{x^3}{x} = x^2).
2. Multiply $(x - 2)$ by $x^2$ to get $x^3 - 2x^2$.
3. Subtract: $P(x) - (x^3 - 2x^2) = 5x^2 - 13x + 6$.
4. Repeat the process:
   • Divide: (\frac{5x^2}{x} = 5x).
   • Multiply: $5x(x - 2) = 5x^2 - 10x$.
   • Subtract: $5x^2 - 13x + 6 - (5x^2 - 10x) = -3x + 6$.
5. Repeat again:
   • Divide: (\frac{-3x}{x} = -3).
   • Multiply: $-3(x - 2) = -3x + 6$.
   • Subtract: $-3x + 6 - (-3x + 6) = 0$.

Therefore, we have:

$$P(x) = (x - 2)(x^2 + 5x - 3).$$

Thus, we can factor $P(x)$ as $A(x) = (x - 2)$ and $B(x) = (x^2 + 5x - 3)$.

Two vectors are perpendicular if their dot product equals zero:

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

This gives:

$$a(2a - 3) - 2 = 0.$$

Expanding it:

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

Using the quadratic formula:

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

This simplifies to:

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

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

The graph of $y = f(x)$ is a quadratic opening upwards with vertex at its minimum point. Since $f(x)$ will be positive for $x < -3$ and $x > 0$, the corresponding values on the graph of $y = \frac{1}{f(x)}$ will be low in those regions.

However, as $f(x)$ approaches the vertex minimum, rac{1}{f(x)} will tend toward infinity.

To sketch:

1. Identify vertical asymptotes near the minimum of $f(x)$ (at $x = -1$ where $f(-1)$ is a minimum).
2. Sketch curve approaching these asymptotes from above and below.
3. Ensure the curve reflects typical characteristics of rational functions.