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 verify that x - 4 is a factor, we use polynomial long division or synthetic division.
Substituting x = 4 into f:
f(4) = (4)^3 - 2(4)^2 - 7(4) - 4 = 64 - 32 - 28 - 4 = 0.
Since f(4) = 0, x - 4 is indeed a factor of f.

To find the x-intercepts, we set f(x) = 0:
$x^3 - 2x^2 - 7x - 4 = 0$.
Factoring gives: $(x - 4)(x^2 + 2x + 1) = 0$.
Solving, we find:

1. $x - 4 = 0 o x = 4$,
2. $x^2 + 2x + 1 = 0$, which factors to $(x + 1)^2 = 0 o x = -1$.
   Thus, the x-intercepts are (4, 0) and (-1, 0).

To find the turning points, we first find the derivative:
f'(x) = 3x^2 - 4x - 7.
Setting f'(x) = 0:
$3x^2 - 4x - 7 = 0$.
Using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4(3)(-7)}}{2(3)} = \frac{4 \pm \sqrt{16 + 84}}{6} = \frac{4 \pm 10}{6}$.
This results in:

1. $x = \frac{14}{6} = \frac{7}{3}$,
2. $x = \frac{-6}{6} = -1$.
   Evaluating f at these points gives the coordinates:
3. For $x = \frac{7}{3}$,
   f(\frac{7}{3}) = \frac{500}{27} o \left(\frac{7}{3}, \frac{500}{27}\right)$,
4. For $x = -1$,
   f(-1) = -4 o (-1, -4).
   Thus, the turning points are \left(\frac{7}{3}, \frac{500}{27}\right) \text{ and } (-1, -4).

To sketch the graph, plot the y-intercept (0, -4), x-intercepts (4, 0) and (-1, 0), and turning points \left(\frac{7}{3}, \frac{500}{27}\right) and (-1, -4).
Ensure the curve reflects the shapes between these points and includes the appropriate behavior as x approaches positive and negative infinity.

Since the graph is decreasing where (f'(x) < 0),
set the derivative:
$3x^2 - 4x - 7 < 0$.
Using the solutions found earlier, analyze intervals around $x = -1$ and $x = \frac{7}{3}$.
The graph is decreasing in the interval: $(-\infty, -1)$ and $(\frac{7}{3}, \infty)$.