The diagram shows the graph of the cubic function $f$ defined for $x \in \mathbb{R}$ as follows: $$ f : x \mapsto x^3 - x^2 - x + 6 $$ (a) Find the co-ordinates of the point at which $f$ cuts the $y$-axis - Leaving Cert Mathematics - Question Question 1 - 2012

Given that $f$ has a minimum turning point at $(1, 5)$, we evaluate the derivative:

$f'(x) = 3x^2 - 2x - 1.$

Setting $f'(x) = 0$, we get:

$3x^2 - 2x - 1 = 0.$

Using the quadratic formula, we find:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm 4}{6},$

which gives us:

$x = 1 \quad \text{or} \quad x = -\frac{1}{3}.$

To find the corresponding $y$ value:

$f\left(-\frac{1}{3}\right) = \left(-\frac{1}{3}\right)^3 - \left(-\frac{1}{3}\right)^2 - \left(-\frac{1}{3}\right) + 6 = -\frac{1}{27} - \frac{1}{9} + \frac{1}{3} + 6 = \frac{-1 -3 + 9 + 162}{27} = \frac{167}{27}.$

Thus, the maximum turning point is at $\left(-\frac{1}{3}, \frac{167}{27}\right)$.

The slope of the line $k$ is given as:

$k : 4x - y + 9 = 0 \implies y = 4x + 9.$

Thus, the slope of $l$ is 4.

Next, we find the derivative of the function:

$f'(x) = 3x^2 - 2x - 1.$

Setting the derivative equal to the slope:

$3x^2 - 2x - 1 = 4 \implies 3x^2 - 2x - 5 = 0.$

Using the quadratic formula again:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3} = \frac{2 \pm \sqrt{4 + 60}}{6} = \frac{2 \pm 8}{6}.$

This gives:

$x = \frac{10}{6} = \frac{5}{3} \quad \text{or} \quad x = -1.$

Thus, the $x$-coordinate at which $l$ is tangent to the curve is at $x = \frac{5}{3}$.