Given: $f(x) = x^3 - 2x^2 - 7x - 4$ 7.1 Write down the $y$-intercept of $f$ - NSC Technical Mathematics - Question 7 - 2021 - Paper 1

To show that $x - 4$ is a factor, we can substitute $x = 4$ into the function:

$f(4) = (4)^3 - 2(4)^2 - 7(4) - 4$
Calculating, we get:
$f(4) = 64 - 32 - 28 - 4 = 0$
Since $f(4) = 0$, it indicates that $x - 4$ is a factor of $f$.

To find the $x$-intercepts, we set $f(x)$ to zero:

$x^3 - 2x^2 - 7x - 4 = 0$
Using synthetic division with $x - 4$:

1. First factor: $(x - 4)(x^2 + 2x + 1) = 0$
2. The quadratic factors as: $(x + 1)^2 = 0$
   Thus, the $x$-intercepts are $x = 4$ and $x = -1$ (multiplicity 2).

To find turning points, we first compute the derivative:

$f'(x) = 3x^2 - 4x - 7$
Setting the derivative to zero:

$3x^2 - 4x - 7 = 0$
Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4(3)(-7)}}{2(3)}$
Calculating gives:
$x = \frac{4 \pm \sqrt{16 + 84}}{6} = \frac{4 \pm \sqrt{100}}{6} = \frac{4 \pm 10}{6}$
Thus,

1. x = rac{14}{6} = \frac{7}{3}
2. $x = -1$
   Next, substituting back into $f(x)$:
3. For $x = \frac{7}{3}$:
   $f(\frac{7}{3}) = ...$
   Calculation will yield the $y$-coordinate as rac{500}{27} (approx. 18.52).
4. For $x = -1$:
   $f(-1) = -1 + 2 + 7 - 4 = 4$
   So, turning points are ((\frac{7}{3}, \frac{500}{27})) and $(-1, 4)$.

The graph of $f$ is a cubic function, and it will have an upward and downward shape, reflecting the nature of its turning points found earlier. Remember to highlight the following features:

• $y$-intercept at $(0, -4)$
• $x$-intercepts at $(-1, 0)$ and $(4, 0)$
• Turning points at $(\frac{7}{3}, \frac{500}{27})$ and $(-1, 4)$.

The graph of $f$ is decreasing where the derivative is negative:

1. Find critical points: from part 7.4, we have $x = -1$ and $x = \frac{7}{3}$.
2. Analyzing:

• For $(-\infty, -1)$, choose $x = -2$, $f'(-2) > 0$
• For $(-1, \frac{7}{3})$, choose $x = 0$, $f'(0) < 0$
• For $(\frac{7}{3}, \infty)$, choose $x = 3$, $f'(3) > 0$
  Thus, $f$ is decreasing in the interval $(-1, \frac{7}{3})$.