Question2. The curve shown is part of the graph of $y=x^{3}-x^{2}-x+3$. The point A is a local maximum and the point B is a point of inflexion. (a) (i) Find the coordinates of A. (ii) Find the coordinates of B. (b) (i) Find the equation of the line containing both A and B. (ii) Find the equation of a tangent to the curve which is parallel to this line. [10 marks]

Studdy Solution
Find the equation of a tangent to the curve which is parallel to this line. The slope of the tangent must be $-\frac{8}{9}$.
Set the derivative equal to the slope:
$$3x^2 - 2x - 1 = -\frac{8}{9}$$
Multiply through by 9 to clear the fraction:
$$27x^2 - 18x - 9 = -8$$
$$27x^2 - 18x - 1 = 0$$
Solve this quadratic equation using the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where $a = 27$, $b = -18$, and $c = -1$.
$$x = \frac{-(-18) \pm \sqrt{(-18)^2 - 4 \cdot 27 \cdot (-1)}}{2 \cdot 27}$$
$$x = \frac{18 \pm \sqrt{324 + 108}}{54}$$
$$x = \frac{18 \pm \sqrt{432}}{54}$$
$$x = \frac{18 \pm 12\sqrt{3}}{54}$$
$$x = \frac{3 \pm 2\sqrt{3}}{9}$$
$$x = \frac{1}{3} \pm \frac{2\sqrt{3}}{9}$$
Calculate the y-coordinates for these x-values and find the tangent equations using point-slope form.
The equations of the tangents are:
$$y = -\frac{8}{9}x + \text{constant}$$