To find the point \((x, y)\) where the tangent to the curve \( y = x^3 \) at \( x = 2 \) intersects the curve itself, follow these steps:

1. **Find the coordinates of the point on the curve:**
   \[
   y = x^3 \quad \text{at } x = 2
   \]
   \[
   y = 2^3 = 8
   \]
   So, the point on the curve is \((2, 8)\).

2. **Find the slope of the tangent line:**
   The slope of the tangent line at any point on the curve is given by the derivative of \( y = x^3 \). The derivative is:
   \[
   \frac{dy}{dx} = 3x^2
   \]
   At \( x = 2 \), the slope is:
   \[
   \frac{dy}{dx} \bigg|_{x=2} = 3 \times 2^2 = 12
   \]

3. **Find the equation of the tangent line:**
   Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency:
   \[
   y - 8 = 12(x - 2)
   \]
   Simplifying, we get:
   \[
   y = 12x - 16
   \]

4. **Find the x and y where the tangent intersects the curve again:**
   To find where the tangent line intersects the curve \( y = x^3 \) again, set the equation of the line equal to the equation of the curve:
   \[
   12x - 16 = x^3
   \]
   Rearrange to form a cubic equation:
   \[
   x^3 - 12x + 16 = 0
   \]
   We know \( x = 2 \) is a root since it is the point of tangency. Factoring \( (x - 2) \) from the cubic equation:

   \[
   x^3 - 12x + 16 = (x - 2)(x^2 + 2x - 8) = 0
   \]

   Solve \( x^2 + 2x - 8 = 0 \) using the quadratic formula:

   \[
   x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-8)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2}
   \]
   \[
   x = \frac{-2 \pm 6}{2}
   \]

   So, the solutions are:

   \[
   x = 2 \quad \text{and} \quad x = -4
   \]

   The intersection point other than \( x = 2 \) is \( x = -4 \).

5. **Find the corresponding \( y \)-coordinate for \( x = -4 \):**

   Substitute \( x = -4 \) back into the equation of the curve \( y = x^3 \):

   \[
   y = (-4)^3 = -64
   \]

   So, the tangent intersects the curve again at \((-4, -64)\).

In summary:

- **Point of tangency:** \((2, 8)\)
- **Other intersection point:** \((-4, -64)\)

Would you like further details or have any questions?

Here are 5 related questions:

1. How do you factor a cubic polynomial when given a root?
2. What is the geometric significance of a tangent line intersecting a curve at more than one point?
3. How can you find the points of inflection of the curve \(y = x^3\)?
4. What is the general formula for finding the equation of a tangent to any function \(y = f(x)\)?
5. How do we use derivatives to find concavity and points of inflection?

**Tip:** Whenever you're dealing with tangents, the derivative of the function gives you the slope at any point on the curve.

Learn how to find the tangent to the curve y = x^3 at x = 2 and where the tangent intersects the curve again. This solution involves differentiation, solving cubic equations, and applying the quadratic formula. Suitable for high school students in Grades 10-12.

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