Given: $f(x) = (x - 1)^2(x + 3)$ 9.1 Determine the turning points of $f$ - NSC Mathematics - Question 9 - 2017 - Paper 1

The $y$-intercept can be found by evaluating $f(0)$:

$f(0) = (0 - 1)^2(0 + 3) = 1^2 \cdot 3 = 3$

The $x$-intercepts are obtained by solving $f(x) = 0$:

$(x - 1)^2(x + 3) = 0$

Setting each factor to zero gives us $x = 1$ (double root) and $x = -3$. The turning points are non-existent here, thus the function is continuous without local extrema. Sketching these points results in a cubic shape traversing from the third quadrant up through the first quadrant.

To find the point where concavity changes, we compute the second derivative:

$f''(x) = 6x - 5$

Setting $f''(x) = 0$ gives us:

$6x - 5 = 0 \Rightarrow x = \frac{5}{6}$

Now calculate $f(\frac{5}{6})$:

$f(\frac{5}{6}) = (\frac{5}{6} - 1)^2(\frac{5}{6} + 3) = (\frac{-1}{6})^2(\frac{23}{6}) = \frac{1}{36} \cdot \frac{23}{6} = \frac{23}{216}$

Thus the coordinates are $\left(\frac{5}{6}, \frac{23}{216}\right)$.

For $f(x) = k$ to have three distinct roots, $k$ must be less than the maximum value of $f$. The maximum occurs at $x = \frac{5}{6}$:

$f\left(\frac{5}{6}\right) = \frac{23}{216}$

Thus, the condition is:

$k < \frac{23}{216}$

The slope of the given line is $-5$. We set $f'(x) = -5$:

$3x^2 - 5x + 3 = -5$

This simplifies to:

$3x^2 - 5x + 8 = 0$

Using the quadratic formula:

$x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 3 \cdot 8}}{2 \cdot 3} = \frac{5 \pm \sqrt{25 - 96}}{6}$

This results in no real roots, meaning that there's no point of tangency on the domain $x < 0$.