At what point, the slope of the curve $y = - {x^3} + 3{x^2} + 9x - 27$ is maximum? Also, find the maximum slope.

We have, $y = - {x^3} + 3{x^2} + 9x - 27$

Therefore,$\frac{{dy}}{{dx}} = - 3{x^2} + 6x + 9 =$ Slope of tangent to the curve

Now, $\frac{{{d^2}y}}{{d{x^2}}} = - 6x + 6$

For $\frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = 0$,
$- 6x + 6 = 0$

$\Rightarrow$ $x = \frac{{ - 6}}{{ - 6}} = 1$

Therefore, $\frac{d}{{dx}}\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right) = - 6 < 0$

So, the slope of tangent to the curve is maximum, when $x = 1$.

For $x = 1$, ${\left( {\frac{{dy}}{{dx}}} \right)_{(x = 1)}} = - 3 \cdot {1^2} + 6 \cdot 1 + 9 = 12$,

which is maximum slope.

Also, for $x = 1,y = - {1^3} + 3 \cdot {1^2} + 9 \cdot 1 - 27$
$= - 1 + 3 + 9 - 27$
$= - 16$

Therefore the required point is (1,-16).