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

After verifying that $x - 2$ is a factor, we can perform polynomial long division to factor $P(x)$.

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

1. Divide the leading term: rac{x^3}{x} = x^2.

2. Multiply back: $x^2(x - 2) = x^3 - 2x^2.$

3. Subtract:
   $P(x) - (x^3 - 2x^2) = 5x^2 - 13x + 6.$

4. Repeat this process until we factor completely.

Ultimately, we have: $P(x) = (x - 2)(x^2 + 5x - 3).$

Two vectors are perpendicular if their dot product is zero. We compute the dot product: egin{pmatrix} a \ -1 \ \frac{2a - 3}{2} \end{pmatrix} \cdot \begin{pmatrix} 2a - 3 \ 2 \end{pmatrix} = 0.

This expands to: $a(2a - 3) + (-1)(2) + \frac{(2a - 3)(2)}{2} = 0.$

Solving this equation yields: $0 = 2a^2 - 3a - 2 + 2a - 3 = 2a^2 - a - 5 = 0.$

Applying the quadratic formula: $a = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-5)}}{2(2)} = \frac{1 \pm 5}{4}$

Thus, $a = \frac{3}{2}$ or $a = -1$.

To sketch the graph of $y = \frac{1}{f(x)}$, we first identify key features of $f(x)$. The graph touches the x-axis at $x=2$ and opens upwards, reaching its local minimum.

When $f(x) = 0$ (i.e., at $x = 2$), the graph of $y = \frac{1}{f(x)}$ will exhibit a vertical asymptote.

Moreover, for large positive or negative values of $x$, $f(x)$ approaches infinity, causing $y = \frac{1}{f(x)}$ to approach zero.

Hence, the sketch will reflect these features, indicating the behavior of $f(x)$ and its key points in relation to the asymptotic behavior of the reciprocal function.