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

The function has intercepts at:

• x-intercepts: Set $f(x) = 0$.

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

This gives $x = 1$ (with a multiplicity of 2) and $x = -3$.

• y-intercept: Set $x = 0$.

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

The sketch should reflect these points and show that the graph touches the x-axis at $x = 1$ and crosses at $x = -3$.

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

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

Setting the second derivative equal to zero:

$6x - 5 = 0$

This gives us:

$x = \frac{5}{6}$

Now we substitute to find $f\left(\frac{5}{6}\right)$:

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

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

For the equation $f(x) = k$ to have three distinct roots, the horizontal line $y = k$ must intersect the graph of $f$ in three places. Given the turning points and the shape of the function, we observe that:

• The maximum value of $f(x)$ occurs at the vertex of the parabola, which can be calculated as:

$f(\frac{1}{3}) = \left(\frac{1}{3} - 1\right)^2\left(\frac{1}{3} + 3\right) = \left(-\frac{2}{3}\right)^2\left(\frac{10}{3}\right) = \frac{4}{9} \cdot \frac{10}{3} = \frac{40}{27}$

Thus, $k$ must satisfy:

$0 < k < \frac{40}{27}$

The line $y = -5x$ has a slope of $-5$. To find the point on $f$ where the tangent is also $-5$, we first find where:

$f'(x) = -5$

We already calculated:

$f'(x) = 3x^2 - 5x + 3$

Setting the slope equal to $-5$ gives:

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

This simplifies to:

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

Using the quadratic formula:

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

This will result in no real solutions, indicating no tangent parallel to $y = -5x$ exists for $x < 0$.