A) i) Draw the graph of f(x) = x² − x − 6 and hence use it to solve the equation x² − x − 12 = 0.

This is a parabola opening upwards since the coefficient of $x^2$ is positive (1).

• Vertex: The x-coordinate of the vertex is $x = -\frac{B}{2A} = -\frac{-1}{2(1)} = \frac{1}{2}$. The y-coordinate of the vertex is $f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) - 6 = \frac{1}{4} - \frac{2}{4} - \frac{24}{4} = -\frac{25}{4} = -6.25$. The vertex is $\left(\frac{1}{2}, -\frac{25}{4}\right)$.

• x-intercepts (roots): Set $f(x) = 0$.

  $$x^2 - x - 6 = 0$$ $$(x-3)(x+2) = 0$$ $$x = 3 \quad \text{or} \quad x = -2$$

  The x-intercepts are $(3,0)$ and $(-2,0)$.

• y-intercept: Set $x = 0$.

  $$f(0) = (0)^2 - (0) - 6 = -6$$

  The y-intercept is $(0,-6)$.

1. Plot the vertex at $\left(\frac{1}{2}, -\frac{25}{4}\right)$ or $(0.5, -6.25)$.
2. Plot the x-intercepts at $(-2,0)$ and $(3,0)$.
3. Plot the y-intercept at $(0,-6)$.
4. Draw a smooth parabola passing through these points, opening upwards.

We want to solve $x^2 - x - 12 = 0$.
We know $f(x) = x^2 - x - 6$.
We can rewrite the equation as:

To solve $x^2 - x - 12 = 0$ graphically, we need to find the x-values where the graph of $y = f(x)$ intersects the horizontal line $y=6$.

Draw the horizontal line $y=6$ on the same coordinate plane as $f(x)$.
The x-coordinates of the points where the parabola $y = x^2 - x - 6$ intersects the line $y=6$ are the solutions.
To find these points algebraically for verification:

The solutions to the equation $x^2 - x - 12 = 0$ are the x-coordinates of the intersection points, which are $x=-3$ and $x=4$.

The solutions are $\boxed{x=-3, x=4}$.

a) ii) Express the function $f(x) = -5x^2 + 9x - 7$ in the form of $f(x) = -a(x+b)^2 + c$ hence use the result to identify the turning points.

Factor out the coefficient of $x^2$ from the first two terms:

Take half of the coefficient of $x$ ($-\frac{9}{5}$), which is $-\frac{9}{10}$, and square it: $\left(-\frac{9}{10}\right)^2 = \frac{81}{100}$.
Add and subtract this value inside the parenthesis:

Group the perfect square trinomial:

Distribute the $-5$:

Combine the constant terms:

By comparing $f(x) = -5\left(x - \frac{9}{10}\right)^2 - \frac{59}{20}$ with $f(x) = -a(x+b)^2 + c$:
We have $-a = -5 \implies a = 5$.
We have $x+b = x - \frac{9}{10} \implies b = -\frac{9}{10}$.
We have $c = -\frac{59}{20}$.

The function in