#### Maximum slope of the curve $y = -x^3 + 3x^2 + 9x - 27$ is: A. 0 B. 12 C. 16 D. 32

Given equation of curve is $y = -x^3 + 3x^2 + 9x - 27$

Apply first derivative and get
$\frac{d y}{d x}=\frac{d\left(-x^{3}+3 x^{2}+9 x-27\right)}{d x}$
Apply sum rule and $\varrho$ is the differentiation of the constant term, so
$\frac{d y}{d x}=-\frac{d\left(x^{3}\right)}{d x}+3 \frac{d\left(x^{2}\right)}{d x}+9 \frac{d(x)}{d x}-0$
Apply power rule and get
$\frac{dy}{dx} = -3x^2 + 3 \cdot (2x) + 9.1 \\ \frac{dy}{dx} = -3x^2 + 6x + 9 \quad \ldots \ldots \ldots \text{(i)}$

Hence, it is the slope of the curve.

Now to find out the second derivative of the given curve, we will differentiate equation (i) once again

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}\left(-3 \mathrm{x}^{2}+6 \mathrm{x}+9\right)}{\mathrm{dx}}$
Apply sum rule and 0 is the differentiation of the constant term so
$\frac{d^{2} y}{d x^{2}}=-3 \frac{d\left(x^{2}\right)}{d x}+6 \frac{d(x)}{d x}+0$
Apply power rule and get
$\frac{d^2 y}{dx^2} = -3(2x) + 6.1 \\ \frac{d^2 y}{dx^2} = -6x + 6 = -6(x - 1) \ldots \text{(ii)}$

Now we will find the critical point by equating the second derivative to 0, we get

-6(x-1) =0

⇒ x-1=0

⇒ x=1

Now, to find out the third derivative of the given curve, we will differentiate equation (ii) once again

$\frac{d^{3} y}{d x^{3}}=\frac{d(-6 x+6)}{d x}$
Apply sum rule and 0 is the differentiation of the constant term, so
$\frac{d^{3} y}{d x^{3}}=-6 \frac{d(x)}{d x}+0$
Apply power rule and get
$\frac{d^{3} y}{d x^{3}}=-6<0$$\frac{d^{3} y}{d x^{3}}=-6<0$
Hence, maximum slope is at $\mathrm{x}=1$
Now, substitute $\mathrm{x}=1$ in (i), and get
$\\ \left(\frac{d y}{d x}\right)_{x=1}=-3 x^{2}+6 x+9 \\\left(\frac{d y}{d x}\right)_{x=1}=-3(1)^{2}+6(1)+9=-3+6+9=12$

Therefore, 12 is the maximum slope of the curve $y = -x^3 + 3x^2 + 9x - 27$.

So, the correct answer is option B.