Question

The curve y= x^3 + 8 intersects the x-axis at C. The tangent to the curve at C intersects the y-axis at D. Find the distance CD?

Answer 1

To solve this problem, we first need to find the point of intersection \( C \) where the curve \( y = x^3 + 8 \) intersects the x-axis. At the x-axis, \( y = 0 \).

1. **Finding point C:**
\[
0 = x^3 + 8
\]
\[
x^3 = -8
\]
\[
x = -2
\]
Therefore, point \( C = (-2, 0) \).

2. **Finding the derivative of the curve:**
We need to find the slope of the tangent at point \( C \).
\[
y' = \frac{d}{dx}(x^3 + 8) = 3x^2
\]
Evaluating the derivative at \( x = -2 \):
\[
y'(-2) = 3(-2)^2 = 3 \cdot 4 = 12
\]
Thus, the slope of the tangent at point \( C \) is 12.

3. **Finding the equation of the tangent line at point C:**
The equation of the tangent line can be written in point-slope form:
\[
y - y_1 = m(x - x_1)
\]
where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency \( C(-2, 0) \).
Substituting the values:
\[
y - 0 = 12(x + 2)
\]
Simplifying this:
\[
y = 12x + 24
\]

4. **Finding the y-intercept (point D) of the tangent line:**
The y-intercept occurs when \( x = 0 \):
\[
y = 12(0) + 24 = 24
\]
Thus, the coordinates of point \( D \) are \( (0, 24) \).

5. **Finding the distance \( CD \):**
We can find distance \( CD \) using the distance formula:
\[
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
where \( (x_1, y_1) = (-2, 0) \) and \( (x_2, y_2) = (0, 24) \).
Substituting the coordinates:
\[
CD = \sqrt{(0 - (-2))^2 + (24 - 0)^2} = \sqrt{(2)^2 + (24)^2}
\]
Calculating:
\[
CD = \sqrt{4 + 576} = \sqrt{580} = \sqrt{4 \cdot 145} = 2 \sqrt{145}
\]

Thus, the distance \( CD \) is \( \boxed{2\sqrt{145}} \).