Given a function defined by $f(x) = -x^3 + 6x^2 - 3x - 10 = -(x - 2)(x^2 - 4x - 5)$ 7.1 Write down the coordinates of the y-intercept of $f$ - NSC Technical Mathematics - Question 7 - 2023 - Paper 1

To show that $(x + 1)$ is a factor, we evaluate $f(-1)$:

$f(-1) = -(-1)^3 + 6(-1)^2 - 3(-1) - 10 = 1 + 6 + 3 - 10 = 0.$
Since $f(-1) = 0$, $(x + 1)$ is a factor of $f$.

To find the x-intercepts, we solve for $f(x) = 0$:

Using the factorization $f(x) = -(x - 2)(x + 1)(x - 5)$, we get:

1. $x - 2 = 0 ightarrow x = 2$
2. $x + 1 = 0 ightarrow x = -1$
3. $x - 5 = 0 ightarrow x = 5$
   Therefore, the x-intercepts are $(-1, 0)$, $(2, 0)$, and $(5, 0)$.

First, find the derivative of $f$:

$f'(x) = -3x^2 + 12x - 3.$
Setting the derivative to zero:

$-3x^2 + 12x - 3 = 0$
Dividing through by -3 gives:
$x^2 - 4x + 1 = 0.$
Using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}.$
To find the y-coordinates, substitute back into $f$:
The turning points are therefore approximately $(0.27, -10)$ and $(3.73, 10.39)$.

The graph of $f$ is a cubic function with the following features:

• The y-intercept is at $(0, -10)$.
• The x-intercepts are at $(-1, 0)$, $(2, 0)$, and $(5, 0)$.
• The turning points are approximately at $(0.27, -10)$ and $(3.73, 10.39)$, showing the local maximum and minimum.
  The graph has the typical cubic shape, showing smooth curves.

From the graph, identify the intervals where $f'(x) > 0$.
Since $f'(x)$ is positive when the graph is increasing, the valid values of $x$ that satisfy $x > 0$ and $x \times f'(x) > 0$ are approximately in the interval $(0.27, 3.73)$.