The graph below represents the function defined by $f(x) = x^3 + 3x^2 - 9x + k$ and cuts the x-axis at A $(1; 0)$ and B - NSC Technical Mathematics - Question 7 - 2022 - Paper 1

To determine the value of $k$, we substitute point A $(1;0)$ into the function:

$f(1) = 1^3 + 3(1)^2 - 9(1) + k = 0$
This simplifies to:
$1 + 3 - 9 + k = 0$
$k - 5 = 0$
Thus, $k = 5$.

To find point B, we need to find the x-intercepts of the function $f(x) = x^3 + 3x^2 - 9x + 5$.
Setting the function to zero: $f(x) = 0 \Rightarrow (x - 1)(x + 5)(x - 1) = 0$
Thus, the x-coordinates of point B are $(-5; 0)$.

To find the turning points, we first differentiate the function:
$f'(x) = 3x^2 + 6x - 9$
Setting the derivative to zero:
$3(x + 3)(x - 1) = 0$
The solutions are $x = -3$ and $x = 1$.
At $x = -3$, we substitute back into the original function:
$f(-3) = (-3)^3 + 3(-3)^2 - 9(-3) + 5 = 32$
Thus, the coordinates of turning point D are $(-3; 32)$.

The critical values occur at $x = -3$ and $x = 1$.
Therefore, $x$ lies within the interval:
$x \in [-3; 1]$

The new coordinates of point A when shifted down by 2 units are $(1; 0 - 2) = (1; -2)$.