The slope of the tangent line to the curve y=2x3 at the point (−1,−2) is:

slope = 

The equation of this tangent line can be written in the form y=mx+b where

m= 

and where

b= 

**Note**: *You can earn partial credit on this problem.*

To find the slope of the tangent line to the curve \( y = 2x^3 \) at the point \( (-1, -2) \), follow these steps:

### Step 1: Find the derivative (slope function) of \( y = 2x^3 \)

The slope of the tangent line at any point is given by the derivative of the function. Let's differentiate \( y = 2x^3 \) with respect to \( x \).

\[
\frac{dy}{dx} = \frac{d}{dx}(2x^3) = 6x^2
\]

### Step 2: Evaluate the derivative at the point \( (-1, -2) \)

Now, substitute \( x = -1 \) into the derivative to find the slope of the tangent line at this point:

\[
\text{slope} = 6(-1)^2 = 6(1) = 6
\]

So, the slope \( m = 6 \).

### Step 3: Find the equation of the tangent line

The equation of the tangent line is in the form \( y = mx + b \). We know \( m = 6 \) and the point \( (-1, -2) \) lies on the tangent line. Use the point-slope form to find \( b \):

\[
y = mx + b
\]

Substitute \( m = 6 \), \( x = -1 \), and \( y = -2 \):

\[
-2 = 6(-1) + b
\]

\[
-2 = -6 + b
\]

Solve for \( b \):

\[
b = -2 + 6 = 4
\]

### Final Answer:

The slope of the tangent line is:

\[
\text{slope} = 6
\]

The equation of the tangent line is:

\[
y = 6x + 4
\]

So, \( m = 6 \) and \( b = 4 \).

Would you like more details or have any questions?

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Here are 5 questions to expand your understanding:
1. How is the derivative related to the slope of the tangent line at a point?
2. What is the general process for finding the equation of a tangent line?
3. How would the tangent line change if the point of tangency was different?
4. What role does the constant \( b \) play in the equation of a line?
5. What happens if the slope at a point is zero? What does that imply about the tangent line?

**Tip:** The derivative gives you the rate of change at a specific point. For curves, this is the slope of the tangent line.

The slope of the tangent line to the curve y=2x^3 at the point (-1,-2) is: slope = The equation of this tangent line can be written in the form y=mx+b where m= and where b=

Learn how to find the slope and equation of the tangent line to the curve y=2x^3 at the point (-1,-2). This calculus problem covers derivative concepts and the point-slope form of a line, suitable for Grades 11-12 students.

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