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To solve this problem, we need to find both the equation of the tangent line and the normal line to the function \( y = x^3 \) at \( x = 2 \).

### Step 1: Find the derivative
The derivative of the function gives us the slope of the tangent line at any point. Let's differentiate \( y = x^3 \):

\[
\frac{dy}{dx} = 3x^2
\]

### Step 2: Find the slope of the tangent line at \( x = 2 \)
Now, substitute \( x = 2 \) into the derivative to find the slope at that point:

\[
\frac{dy}{dx}\Bigg|_{x=2} = 3(2)^2 = 12
\]

So, the slope of the tangent line at \( x = 2 \) is 12.

### Step 3: Find the coordinates of the point of tangency
We need the coordinates of the point on the curve where \( x = 2 \). Substitute \( x = 2 \) into the original function \( y = x^3 \):

\[
y = (2)^3 = 8
\]

Thus, the point of tangency is \( (2, 8) \).

### Step 4: Equation of the tangent line
The equation of a line is given by the formula:

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

where \( m \) is the slope, and \( (x_1, y_1) \) is the point of tangency. Substituting \( m = 12 \) and \( (x_1, y_1) = (2, 8) \):

\[
y - 8 = 12(x - 2)
\]

Simplifying:

\[
y - 8 = 12x - 24
\]

\[
y = 12x - 16
\]

So, the equation of the tangent line is:

\[
y = 12x - 16
\]

### Step 5: Find the equation of the normal line
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line is 12, the slope of the normal line is:

\[
m_{\text{normal}} = -\frac{1}{12}
\]

Using the same point \( (2, 8) \) and the point-slope formula:

\[
y - 8 = -\frac{1}{12}(x - 2)
\]

Simplifying:

\[
y - 8 = -\frac{1}{12}x + \frac{1}{6}
\]

\[
y = -\frac{1}{12}x + \frac{1}{6} + 8
\]

\[
y = -\frac{1}{12}x + \frac{49}{6}
\]

Thus, the equation of the normal line is:

\[
y = -\frac{1}{12}x + \frac{49}{6}
\]

### Summary:
- **Tangent line**: \( y = 12x - 16 \)
- **Normal line**: \( y = -\frac{1}{12}x + \frac{49}{6} \)

Would you like further details or have any other questions?

Here are 5 related questions to expand on this topic:
1. How do we determine the derivative of more complex functions?
2. What is the significance of the negative reciprocal for the normal line's slope?
3. How can we find tangent and normal lines for trigonometric functions?
4. Can we apply this method to implicit functions or parametrically defined curves?
5. What if we wanted to find the tangent and normal lines for functions in three dimensions?

**Tip:** The tangent line represents the instantaneous rate of change of the function at a given point, while the normal line is perpendicular to the tangent and often used in geometry and physics contexts.

Find the equation of the tangent line and normal line to y = x^3 at x = 2.

Learn how to find the equations of the tangent and normal lines to the function y = x^3 at x = 2. This solution involves differentiation to find the slope, applying the point-slope formula, and understanding the relationship between tangent and normal lines. Suitable for students in Grades 11-12.

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