The graphs of $f(x) = x^2 - 2x - 3$ and $g(x) = mx + c$ are drawn below - NSC Mathematics - Question 6 - 2024 - Paper 1

To find the x-intercepts (D and E), set $f(x) = 0$: $x^2 - 2x - 3 = 0$ Factoring gives $(x - 3)(x + 1) = 0$, resulting in the coordinates:

• $D(-1, 0)$
• $E(3, 0)$

The equation of $g$ is determined by finding the slope at point P. At point P where $f(3) = 0$, the slope (gradient) $m$ at point P can be calculated: $f'(x) = 2x - 2 \Rightarrow f'(3) = 4$ Thus, $g(x)$ can be expressed in point-slope form, leading to: $g(x) = 4(x - 3) + 0 \Rightarrow g(x) = 4x - 12$.

The condition $f(x) > g(x)$ means: $x^2 - 2x - 3 > 4x - 12$ Rearranging gives: $x^2 - 6x + 9 > 0$ Factoring gives $(x - 3)^2 > 0$ which is true for all $x \neq 3$.

The vertical distance between $h$ and $g$ is given by: $|h(x) - g(x)| = |-f(x) - g(x)|$ We need to evaluate the maximum over the interval [-2, 3]. Calculating $h(x)$ and $g(x)$ for values within the interval gives the vertical distance, maximizing at: $d = h(-2) - g(-2) = |(4 - 12 - (-12))| = 16$.

For $k(x) = g(x) - n$ to be tangent to $f(x)$, the discriminant must be zero: Setting $g(x) = f(x) + n$: $x^2 - 2x - 3 = 4x - 12 + n$ Rearranging gives: $x^2 - 6x + (n + 9) = 0$ Setting the discriminant $riangle = 0$ yields: $(-6)^2 - 4(1)(n + 9) = 0 \Rightarrow 36 - 4n - 36 = 0 \Rightarrow n = 0$.