a) i) Draw the graph of $f(x) = x^2 - x - 6$ and hence use it to solve the equation $x^2 - x - 12 = 0$. Step 1: Analyze the function $f(x) = x^2 - x - 6$. 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 $(0,-6)$. Step 2: Describe how to draw the graph of $f(x) = x^2 - x - 6$. 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. Step 3: Use the graph to solve $x^2 - x - 12 = 0$. We want to solve $x^2 - x - 12 = 0$. We know $f(x) = x^2 - x - 6$. We can rewrite the equation as: $$x^2 - x - 6 - 6 = 0$$ $$f(x) - 6 = 0$$ $$f(x) = 6$$ 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 $y=6$. Step 4: Find the intersection points. 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: $$x^2 - x - 6 = 6$$ $$x^2 - x - 12 = 0$$ $$(x-4)(x+3) = 0$$ $$x = 4 \quad \text{or} \quad x = -3$$ 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. Step 1: Complete the square for $f(x) = -5x^2 + 9x - 7$. Factor out the coefficient of $x^2$ from the first two terms: $$f(x) = -5\left(x^2 - \frac{9}{5}x\right) - 7$$ 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: $$f(x) = -5\left(x^2 - \frac{9}{5}x + \frac{81}{100} - \frac{81}{100}\right) - 7$$ Group the perfect square trinomial: $$f(x) = -5\left(\left(x - \frac{9}{10}\right)^2 - \frac{81}{100}\right) - 7$$ Distribute the $-5$: $$f(x) = -5\left(x - \frac{9}{10}\right)^2 - 5\left(-\frac{81}{100}\right) - 7$$ $$f(x) = -5\left(x - \frac{9}{10}\right)^2 + \frac{81}{20} - 7$$ Combine the constant terms: $$f(x) = -5\left(x - \frac{9}{10}\right)^2 + \frac{81}{20} - \frac{140}{20}$$ $$f(x) = -5\left(x - \frac{9}{10}\right)^2 - \frac{59}{20}$$ Step 2: Compare with the form $f(x) = -a(x+b)^2 + c$. 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