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The problem is asking for the equation of the tangent line to the curve \( f(x) = 2x^3 + 3x^2 + x \) at the point \( (-1, 0) \).

To find the equation of the tangent line, follow these steps:

### Step 1: Find the derivative of \( f(x) \)
The derivative \( f'(x) \) gives the slope of the tangent line at any point \( x \). Let's differentiate the function:

\[
f(x) = 2x^3 + 3x^2 + x
\]

Using standard differentiation rules:

\[
f'(x) = 6x^2 + 6x + 1
\]

### Step 2: Evaluate the derivative at \( x = -1 \)
Substitute \( x = -1 \) into the derivative to find the slope of the tangent line at that point:

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

So, the slope of the tangent line is \( m = 1 \).

### Step 3: Use the point-slope form of the equation of a line
The point-slope form of a line is:

\[
y - y_1 = m(x - x_1)
\]

Here, \( (x_1, y_1) = (-1, 0) \) and \( m = 1 \). Substituting these values:

\[
y - 0 = 1(x - (-1))
\]

Simplifying:

\[
y = x + 1
\]

Thus, the equation of the tangent line is:

\[
y = x + 1
\]

Would you like more details, or do you have any questions?

### Related questions:
1. How do we find the second derivative of \( f(x) = 2x^3 + 3x^2 + x \)?
2. What is the geometric interpretation of the tangent line to a curve?
3. How would the tangent line equation change if we evaluated it at a different point on the curve?
4. Can we use this method to find the normal line to the curve at the same point?
5. What role does the derivative play in finding the concavity of a function?

**Tip:** The derivative represents the slope of the curve at any given point and is key for understanding rates of change and tangent lines.

Find the equation of the tangent line to the function f(x) = 2x^3 + 3x^2 + x at the point (-1, 0).

Learn how to find the equation of the tangent line for the function f(x) = 2x^3 + 3x^2 + x at the point (-1, 0). This solution covers key calculus concepts, including differentiation and the point-slope formula. Suitable for high school students.

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