What is the maximum slope of the curve `y =-x^(3) + 3x^(2) + 2x - 27` ?

Options

A. 1
B. 2
C. 5
D. `-23`

Answer

To find the maximum slope of the curve given by the equation \( y = -x^3 + 3x^2 + 2x - 27 \), we will follow these steps:

### Step 1: Differentiate the function
To find the slope of the curve, we need to find the first derivative of \( y \) with respect to \( x \).

\[ y' = \frac{dy}{dx} = \frac{d}{dx}(-x^3 + 3x^2 + 2x - 27) \]

Calculating the derivative:

\[ y' = -3x^2 + 6x + 2 \]

### Step 2: Find the second derivative
To determine the maximum slope, we need to find the critical points by setting the second derivative to zero.

\[ y'' = \frac{d^2y}{dx^2} = \frac{d}{dx}(-3x^2 + 6x + 2) \]

Calculating the second derivative:

\[ y'' = -6x + 6 \]

### Step 3: Set the second derivative to zero
Now, we set the second derivative equal to zero to find the critical points.

\[ -6x + 6 = 0 \]

Solving for \( x \):

\[ 6x = 6 \implies x = 1 \]

### Step 4: Find the maximum slope
Now that we have the critical point \( x = 1 \), we will substitute this value back into the first derivative to find the maximum slope.

\[ y'(1) = -3(1)^2 + 6(1) + 2 \]

Calculating this:

\[ y'(1) = -3 + 6 + 2 = 5 \]

### Conclusion
The maximum slope of the curve is \( 5 \).