Given: $f(x) = -x(x - 3)(x - 3)$ 7.1 Write down the coordinates of the x-intercepts of $f$ - NSC Technical Mathematics - Question 7 - 2018 - Paper 1

To find the y-intercept, calculate $f(0)$:
$f(0) = -0(0 - 3)(0 - 3) = 0.$
Thus, the y-intercept is: $(0, 0)$.

Expanding the expression:

1. Expand $(x - 3)(x - 3)$:
   $(x - 3)^2 = x^2 - 6x + 9$
2. Substitute in $f(x)$:
   $f(x) = -x(x^2 - 6x + 9)$
3. Distribute:
   $= -x^3 + 6x^2 - 9x.$
   This shows that $f(x) = -x^3 + 6x^2 - 9x$.

To find the turning points, we calculate the derivative:
$f'(x) = -3x^2 + 12x - 9.$
Setting the derivative to zero:
$-3x^2 + 12x - 9 = 0$
Dividing by -3:
$x^2 - 4x + 3 = 0$
Factoring:
$(x - 3)(x - 1) = 0$
Thus, $x = 3$ and $x = 1$.
Now, to find the coordinates of the turning points, substitute back into $f(x)$:

• For $x = 1$:
  $f(1) = -1^3 + 6(1)^2 - 9(1) = -1 + 6 - 9 = -4.$
• For $x = 3$:
  $f(3) = -3^3 + 6(3)^2 - 9(3) = -27 + 54 - 27 = 0.$
  Thus, the coordinates of the turning points are:
• $(1, -4)$
• $(3, 0)$.

The function is increasing where $f'(x) > 0$.
We previously found:
$f'(x) = -3x^2 + 12x - 9.$
Factoring gives:
$-3(x^2 - 4x + 3) = -3(x - 3)(x - 1) > 0.$
The critical points are $x = 1$ and $x = 3$.
Using test intervals:

• For $x < 1$: $f'(x) > 0$
• For $1 < x < 3$: $f'(x) < 0$
• For $x > 3$: $f'(x) > 0$
  Thus, the intervals where $f$ is increasing are:
• $(- ext{∞}, 1)$ and $(3, ext{∞})$.