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To find the second point where the normal line intersects the parabola, let’s go through the problem step-by-step.

### 1. Find the derivative of \( y = 2 - x + 2x^2 \)
The derivative \( y' \) gives the slope of the tangent line:
\[
y = 2 - x + 2x^2
\]
Taking the derivative with respect to \( x \):
\[
y' = -1 + 4x
\]

### 2. Determine the slope of the tangent line at \( (1, 3) \)
Substitute \( x = 1 \) into \( y' \) to find the slope at this point:
\[
y'(1) = -1 + 4(1) = 3
\]
So, the slope of the tangent line at \( (1, 3) \) is 3.

### 3. Find the slope of the normal line
The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of 3:
\[
\text{slope of normal line} = -\frac{1}{3}
\]

### 4. Write the equation of the normal line
We know the point \( (1, 3) \) lies on the normal line, and the slope is \( -\frac{1}{3} \). Using the point-slope form of a line equation:
\[
y - 3 = -\frac{1}{3}(x - 1)
\]
Expanding and simplifying this equation:
\[
y = -\frac{1}{3}x + \frac{1}{3} + 3
\]
\[
y = -\frac{1}{3}x + \frac{10}{3}
\]

### 5. Find the second intersection point
To find where this normal line intersects the parabola \( y = 2 - x + 2x^2 \) a second time, we substitute \( y = -\frac{1}{3}x + \frac{10}{3} \) into the equation of the parabola:
\[
-\frac{1}{3}x + \frac{10}{3} = 2 - x + 2x^2
\]
Multiply through by 3 to clear the fractions:
\[
-x + 10 = 6 - 3x + 6x^2
\]
Rearrange into standard form:
\[
6x^2 - 2x - 4 = 0
\]
Simplify by dividing by 2:
\[
3x^2 - x - 2 = 0
\]
Now we solve this quadratic equation using the quadratic formula:
\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3}
\]
\[
x = \frac{1 \pm \sqrt{1 + 24}}{6}
\]
\[
x = \frac{1 \pm \sqrt{25}}{6}
\]
\[
x = \frac{1 \pm 5}{6}
\]
This gives us two solutions:
1. \( x = \frac{6}{6} = 1 \) (we already know this point)
2. \( x = \frac{-4}{6} = -\frac{2}{3} \)

### 6. Find the \( y \)-coordinate for the second intersection point
Substitute \( x = -\frac{2}{3} \) into the normal line equation \( y = -\frac{1}{3}x + \frac{10}{3} \):
\[
y = -\frac{1}{3}\left(-\frac{2}{3}\right) + \frac{10}{3}
\]
\[
y = \frac{2}{9} + \frac{10}{3}
\]
Convert \( \frac{10}{3} \) to a fraction with denominator 9:
\[
y = \frac{2}{9} + \frac{30}{9} = \frac{32}{9}
\]

### Final Answer
The second intersection point is:
\[
\left(-\frac{2}{3}, \frac{32}{9}\right)
\]

Would you like more details on any part of this solution? 

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Here are five related questions for further practice:

1. How do you find the equation of a tangent line to a curve at a given point?
2. What are the steps for finding the normal line to a curve at a specific point?
3. How do you determine if two lines are perpendicular based on their slopes?
4. What is the significance of the quadratic formula in finding intersection points?
5. How can we verify if a given point lies on both a line and a curve?

**Tip:** Remember that normal lines are perpendicular to tangent lines, making their slopes negative reciprocals of each other.

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

Learn how to find the second intersection of a line normal to the curve y = 2 - x + 2x^2 at point (1, 3). This solution covers derivatives, slopes of perpendicular lines, and the quadratic formula. Suitable for high school calculus students.

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