Given: Function $f$ defined by $f(x) = -(x - 1)^2 (x + 3) = -x^3 + x^2 + 5x - 3$ 7.1 Write down the $y$-intercept of $f$ - NSC Technical Mathematics - Question 7 - 2020 - Paper 1

To find the $x$-intercepts, set $f(x) = 0$:

$-(x - 1)^2 (x + 3) = 0.$
This gives us two factors to solve:

1. $-(x - 1)^2 = 0$
   • This leads to $x = 1$, with a multiplicity of 2.
2. $(x + 3) = 0$
   • This leads to $x = -3$.
     The $x$-intercepts are therefore $(1, 0)$ and $(-3, 0)$.

Turning points occur where the derivative $f'(x)$ is zero. First, find the derivative of $f$:

$f'(x) = -3x^2 + 2x + 5.$
Set the derivative to zero:

$-3x^2 + 2x + 5 = 0.$
Using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4(-3)(5)}}{-6} = \frac{-2 \pm \sqrt{4 + 60}}{-6} = \frac{-2 \pm \sqrt{64}}{-6} = \frac{-2 \pm 8}{-6}.$
This resolves to:

1. $x = \frac{-10}{-6} = \frac{5}{3}$
2. $x = \frac{6}{-6} = -1$.
   Next, substitute for $y$:
   For $x = \frac{5}{3}$,
   $y = -(\frac{5}{3} - 1)^2 (\frac{5}{3} + 3) = -(\frac{2}{3})^2 (\frac{14}{3}) = -\frac{4}{9} \cdot \frac{14}{3} = -\frac{56}{27}.$
   The first turning point is igg(\frac{5}{3}, -\frac{56}{27}\bigg).
   For $x = -1$:
   $y = -(-1 - 1)^2 (-1 + 3) = -(-2)^2 (2) = -4.$
   The second turning point is $(-1, -4)$.
   Thus, the turning points are igg(\frac{5}{3}, -\frac{56}{27}\bigg) and $(-1, -4)$.

When sketching the graph, plot the points determined earlier:

1. $y$-intercept at $(0, -3)$.
2. $x$-intercepts at $(1, 0)$ and $(-3, 0)$.
3. Turning points at igg(\frac{5}{3}, -\frac{56}{27}\bigg) and $(-1, -4)$.
   Ensure to represent the cubic shape of the function correctly, indicating where the graph crosses the axes and the turning points distinctly.

From the derivative $f'(x) = -3x^2 + 2x + 5$, you set up the inequality:
$-3x^2 + 2x + 5 > 0.$
Factoring or using the quadratic formula gives the critical points. The solutions to the equality correspond to points where the function can change from positive to negative.
Testing intervals around the roots will determine where $f'$ is greater than zero.
The final answer should state the intervals based on the test, which should result in $(-\infty, \frac{5}{3})$ and $(\frac{5}{3}, 1)$ as $f'(x) > 0$ in these regions.