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To solve this problem, let's break it down step by step. The goal is to find the point where the **normal line** to the parabola \( y = 3x + 3x^2 \) at the point \( (1,6) \) intersects the parabola a second time.

### Step 1: Find the slope of the tangent line at \( x = 1 \)
The tangent line to a curve at a point is given by the derivative of the function at that point.

1. The equation of the parabola is:
   \[
   y = 3x + 3x^2
   \]
2. Differentiate \( y \) with respect to \( x \) to find the slope of the tangent line:
   \[
   \frac{dy}{dx} = 3 + 6x
   \]
3. At \( x = 1 \):
   \[
   \frac{dy}{dx} \bigg|_{x=1} = 3 + 6(1) = 9
   \]
   So, the slope of the tangent line at \( (1, 6) \) is 9.

### Step 2: Find the slope of the normal line
The normal line is perpendicular to the tangent line, and the slopes of two perpendicular lines are negative reciprocals. Thus, the slope of the normal line is:
\[
m_{\text{normal}} = -\frac{1}{9}
\]

### Step 3: Find the equation of the normal line
The normal line passes through the point \( (1, 6) \), and its slope is \( -\frac{1}{9} \). Using the point-slope form of a line:
\[
y - y_1 = m(x - x_1)
\]
Substitute \( (x_1, y_1) = (1, 6) \) and \( m = -\frac{1}{9} \):
\[
y - 6 = -\frac{1}{9}(x - 1)
\]
Simplifying:
\[
y - 6 = -\frac{1}{9}x + \frac{1}{9}
\]
\[
y = -\frac{1}{9}x + \frac{55}{9}
\]
This is the equation of the normal line.

### Step 4: Find the point where the normal line intersects the parabola again
To find the second intersection point, substitute the equation of the normal line \( y = -\frac{1}{9}x + \frac{55}{9} \) into the equation of the parabola \( y = 3x + 3x^2 \):
\[
-\frac{1}{9}x + \frac{55}{9} = 3x + 3x^2
\]
Multiply through by 9 to eliminate the fractions:
\[
-x + 55 = 27x + 27x^2
\]
Rearrange the equation:
\[
27x^2 + 28x - 55 = 0
\]

### Step 5: Solve the quadratic equation
We will solve the quadratic equation \( 27x^2 + 28x - 55 = 0 \) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 27 \), \( b = 28 \), and \( c = -55 \). Substituting these values:
\[
x = \frac{-28 \pm \sqrt{28^2 - 4(27)(-55)}}{2(27)}
\]
\[
x = \frac{-28 \pm \sqrt{784 + 5940}}{54}
\]
\[
x = \frac{-28 \pm \sqrt{6724}}{54}
\]
\[
x = \frac{-28 \pm 82}{54}
\]

This gives two solutions for \( x \):
1. \( x = \frac{-28 + 82}{54} = \frac{54}{54} = 1 \) (the first intersection point, which we already know)
2. \( x = \frac{-28 - 82}{54} = \frac{-110}{54} = -\frac{55}{27} \) (the second intersection point)

### Step 6: Find the corresponding \( y \)-value
Substitute \( x = -\frac{55}{27} \) into the equation of the parabola to find the corresponding \( y \)-value:
\[
y = 3\left(-\frac{55}{27}\right) + 3\left(-\frac{55}{27}\right)^2
\]
Simplifying:
\[
y = -\frac{165}{27} + 3 \times \frac{3025}{729}
\]
This simplifies to \( y = 0 \).

### Final Answer:
The second intersection point is \( \left( -\frac{55}{27}, 0 \right) \).

---

Would you like further details or have any questions about this solution?

Here are 5 related questions to explore further:

1. How is the slope of the normal line related to the slope of the tangent line?
2. Why is it necessary to use the quadratic formula in this problem?
3. What would change if the normal line intersected the parabola at a different point?
4. How can we verify that the normal line and parabola only intersect at two points?
5. What are some other real-world applications of normal lines in calculus?

**Tip:** When dealing with intersections between lines and curves, always check if solving the system yields more than

At what point does the normal line to the parabola y = 3x + 3x^2 at (1, 6) intersect the parabola a second time?

Find where the normal line to the parabola y = 3x + 3x² at (1, 6) intersects the curve again. This calculus problem uses differentiation, the concept of perpendicular slopes, and the quadratic formula to find the second point of intersection.

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