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

Substituting the coordinates of point A (1; 0) into the function:

$0 = (1)^3 + 3(1)^2 - 9(1) + k$ $0 = 1 + 3 - 9 + k$ $0 = -5 + k$

Rearranging gives: $k = 5$

Thus, it is shown that k = 5.

We need to find the x-intercepts of the function when $k = 5$:

$f(x) = x^3 + 3x^2 - 9x + 5$ Setting the function to zero to find the coordinates of B: $(x - 1)(x^2 + 4x - 5) = 0$ Solving the quadratic: $(x - 1)(x - 1)(x + 5) = 0$

discriminant method or factoring leads to: $B(-5; 0)$

Thus, the coordinates of point B are (-5, 0).

To find the turning point D, we first need the derivative of the function:

$f’(x) = 3x^2 + 6x - 9$

Setting the derivative to zero to find critical points: $3x^2 + 6x - 9 = 0$

Dividing through by 3 gives: $x^2 + 2x - 3 = 0$ Factoring gives: $(x - 1)(x + 3) = 0$ Thus, $x = -3 \quad \text{or} \quad x = 1$

Using $f(-3)$ to find the coordinates of D:

$f(-3) = (-3)^3 + 3(-3)^2 - 9(-3) + 5 = -27 + 27 + 27 + 5 = 32$

Thus, the coordinates of turning point D are (-3, 32).

From the derivative, we know:

$f’(x) = 3x^2 + 6x - 9$

To find where the function is less than or equal to zero, we first consider the critical values found earlier: $x \\in [-3; 1]$

Here, the output of the derivative is negative between the roots, thus: $x \in [-3; 1].$

When we subtract 2 from the function, $g(x) = f(x) - 2$

The y-coordinate of point A changes:

1. Original A has coordinates (1, 0).
2. The new coordinates are: (1, 0 - 2) = (1, -2).

Thus, the new coordinates of point A are (1, -2).