find the coordinates of the minimum turning point of the curve y=-x3 - 2x2 - x - 2

To find the coordinates of the minimum turning point of the curve \( y = -x^3 - 2x^2 - x - 2 \), we need to follow these steps:

### 1. Differentiate the function
The first step is to find the derivative of \( y \) with respect to \( x \), which will give us the slope of the curve. 

Given:
\[ y = -x^3 - 2x^2 - x - 2 \]

The derivative \( \frac{dy}{dx} \) is:
\[
\frac{dy}{dx} = -3x^2 - 4x - 1
\]

### 2. Set the derivative equal to zero
To find the turning points, we set \( \frac{dy}{dx} = 0 \) and solve for \( x \):

\[
-3x^2 - 4x - 1 = 0
\]

### 3. Solve the quadratic equation
This is a quadratic equation, and we can solve it using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = -3 \), \( b = -4 \), and \( c = -1 \).

Substituting the values:
\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(-3)(-1)}}{2(-3)}
\]
\[
x = \frac{4 \pm \sqrt{16 - 12}}{-6}
\]
\[
x = \frac{4 \pm \sqrt{4}}{-6}
\]
\[
x = \frac{4 \pm 2}{-6}
\]

This gives us two possible solutions for \( x \):
\[
x = \frac{6}{-6} = -1 \quad \text{and} \quad x = \frac{2}{-6} = -\frac{1}{3}
\]

### 4. Determine the nature of the turning points
To determine which of these points is a minimum, we need to find the second derivative \( \frac{d^2y}{dx^2} \):

\[
\frac{d^2y}{dx^2} = -6x - 4
\]

Now, we evaluate \( \frac{d^2y}{dx^2} \) at each turning point:

- For \( x = -1 \):
\[
\frac{d^2y}{dx^2} = -6(-1) - 4 = 6 - 4 = 2
\]

- For \( x = -\frac{1}{3} \):
\[
\frac{d^2y}{dx^2} = -6\left(-\frac{1}{3}\right) - 4 = 2 - 4 = -2
\]

Since \( \frac{d^2y}{dx^2} > 0 \) at \( x = -1 \), this point is a minimum. The other point \( x = -\frac{1}{3} \) is a maximum.

### 5. Find the corresponding \( y \) value
Substitute \( x = -1 \) back into the original equation to find the \( y \)-coordinate of the minimum turning point:

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

### Final Answer:
The coordinates of the minimum turning point are \( (-1, -2) \).

Would you like further details or have any questions?  

Here are some related questions:

1. How do you find the maximum turning point of a cubic function?
2. What is the significance of the second derivative in determining the nature of a turning point?
3. How do you sketch the graph of a cubic function?
4. What are the possible shapes of a cubic function's graph?
5. How do you solve quadratic equations using the quadratic formula?
6. What are inflection points, and how are they different from turning points?
7. Can a cubic function have more than two turning points?
8. How do the coefficients of a cubic equation affect its graph?

**Tip:** When solving quadratic equations, always check if they can be factored before using the quadratic formula, as factoring might provide a quicker solution.

Discover how to find the coordinates of the minimum turning point of the curve y = -x^3 - 2x^2 - x - 2. This problem involves calculus concepts such as derivatives and the second derivative test. Suitable for advanced high school students.

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