# How do you use the first and second derivatives to sketch y= -(x-2) (x+2) (x-4)?

Oct 12, 2016

$y = - \left(x - 2\right) \left(x + 2\right) \left(x - 4\right)$
graph{-(x-2)(x+2)(x-4) [-10, 10, -20, 20]}

#### Explanation:

Fist we find the roots of the equation, ie the values of $x$ st $y = 0$:

$y = 0 \implies - \left(x - 2\right) \left(x + 2\right) \left(x - 4\right) = 0$
$\therefore \left(x - 2\right) \left(x + 2\right) \left(x - 4\right) = 0$
$\therefore x = 2 , x = - 2 , x = 4$

Next we find the critical points, ie the values for which $y ' = 0$, we will need to multiply out as a cubic in order to this:

$y = - \left(x - 2\right) \left(x + 2\right) \left(x - 4\right)$
$\therefore y = - \left(x - 2\right) \left({x}^{2} - 4 x + 2 x - 8\right)$
$\therefore y = - \left(x - 2\right) \left({x}^{2} - 2 x - 8\right)$
$\therefore y = - \left({x}^{3} - 2 {x}^{2} - 8 x - 2 {x}^{2} + 4 x + 16\right)$
$\therefore y = - \left({x}^{3} - 4 {x}^{2} - 4 x + 16\right)$
$\therefore y = - {x}^{3} + 4 {x}^{2} + 4 x - 16$

So, differentiating wrt $x$ gives:
$\therefore y ' = - 3 {x}^{2} + 8 x + 4$
$y ' = 0 \implies - 3 {x}^{2} + 8 x + 4 = 0$
$\therefore 3 {x}^{2} - 8 x - 4 = 0$
$\therefore 3 {x}^{2} - 8 x - 4 = 0$

$x = \frac{- \left(- 8\right) \pm \sqrt{{\left(- 8\right)}^{2} - 4 \left(3\right) \left(- 4\right)}}{2 \left(3\right)}$
$\therefore x = \frac{8 \pm \sqrt{64 + 48}}{6}$
$\therefore x = \frac{8 \pm \sqrt{112}}{6}$
$\therefore x = - 0.43 , 3.10$ (2dp)

, or equally we can solve by completing the square:
$3 {x}^{2} - 8 x - 4 = 0$
$\therefore 3 \left({x}^{2} - \frac{8}{3} x - \frac{4}{3}\right) = 0$
$\therefore \left({x}^{2} - \frac{8}{3} x - \frac{4}{3}\right) = 0$
$\therefore {\left(x - \frac{8}{6}\right)}^{2} - {\left(\frac{8}{6}\right)}^{2} - \frac{4}{3} = 0$
$\therefore {\left(x - \frac{8}{6}\right)}^{2} - \frac{64}{36} - \frac{4}{3} = 0$
$\therefore {\left(x - \frac{8}{6}\right)}^{2} = \frac{64}{36} + \frac{48}{36}$
$\therefore {\left(x - \frac{8}{6}\right)}^{2} = \frac{112}{36}$
$\therefore \left(x - \frac{8}{6}\right) = \pm \sqrt{\frac{112}{36}}$
$\therefore x = \frac{8}{6} \pm \sqrt{\frac{112}{36}}$
again leading to $x = - 0.43 , 3.10$ (2dp)

The coordinates of these points are then given by using the original formula for $y$ to give:
$x = - 0.43 \implies y = - 16.9$ (1dp)
$x = 3.10 \implies y = 5.0$ (1dp)

These values of $x$ correspond to the critical points (or turning points). we now use the second derivative to determine the nature (max, min or point of inflection) of these points.

from earlier, $y ' = - 3 {x}^{2} + 8 x + 4$
$\therefore y ' ' = - 6 x + 8$
$x = - 0.43 \implies y ' ' = 10.5 > 0$ (1dp)
$x = 3.10 \implies y ' ' = - 10.6 < 0$ (1dp)

So we can now determine the nature of the turning points which are:
$\left(- 0.43 , - 16.9\right)$ minimum
$\left(3.10 , 5.0\right)$ maximum

We now have enough to sketch the curve:
graph{-(x-2)(x+2)(x-4) [-10, 10, -20, 20]}