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We are asked to find the second point where the line normal to the parabola \( y = 4x^2 - (3 + 2x) \) at the point \( (1, -1) \) intersects the parabola.

### Step 1: Compute the derivative to find the slope of the tangent line.
The derivative of \( y = 4x^2 - (3 + 2x) \) with respect to \( x \) gives the slope of the tangent line.

\[
\frac{dy}{dx} = \frac{d}{dx} \left( 4x^2 - (3 + 2x) \right)
= 8x - 2
\]

At the point \( (1, -1) \), the slope of the tangent line is:

\[
\frac{dy}{dx} \bigg|_{x=1} = 8(1) - 2 = 6
\]

### Step 2: Find the slope of the normal line.
The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent is 6, the slope \( m_{\text{normal}} \) of the normal line is:

\[
m_{\text{normal}} = -\frac{1}{6}
\]

### Step 3: Equation of the normal line.
The normal line passes through the point \( (1, -1) \) and has a slope of \( -\frac{1}{6} \). Using the point-slope form of the equation of a line:

\[
y - (-1) = -\frac{1}{6}(x - 1)
\]
\[
y + 1 = -\frac{1}{6}(x - 1)
\]
\[
y = -\frac{1}{6}x + \frac{1}{6} - 1
\]
\[
y = -\frac{1}{6}x - \frac{5}{6}
\]

Thus, the equation of the normal line is:

\[
y = -\frac{1}{6}x - \frac{5}{6}
\]

### Step 4: Solve for the second intersection point.
We now need to find the points of intersection between the parabola \( y = 4x^2 - (3 + 2x) \) and the normal line \( y = -\frac{1}{6}x - \frac{5}{6} \). Set the two equations equal to each other:

\[
4x^2 - (3 + 2x) = -\frac{1}{6}x - \frac{5}{6}
\]

Multiply through by 6 to eliminate fractions:

\[
6(4x^2 - 3 - 2x) = -x - 5
\]
\[
24x^2 - 18 - 12x = -x - 5
\]
\[
24x^2 - 12x - 18 = -x - 5
\]
\[
24x^2 - 11x - 13 = 0
\]

### Step 5: Solve the quadratic equation.
We now solve the quadratic equation \( 24x^2 - 11x - 13 = 0 \) using the quadratic formula:

\[
x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(24)(-13)}}{2(24)}
\]
\[
x = \frac{11 \pm \sqrt{121 + 1248}}{48}
\]
\[
x = \frac{11 \pm \sqrt{1369}}{48}
\]
\[
x = \frac{11 \pm 37}{48}
\]

Thus, the two solutions are:

\[
x = \frac{11 + 37}{48} = \frac{48}{48} = 1
\]
\[
x = \frac{11 - 37}{48} = \frac{-26}{48} = -\frac{13}{24}
\]

Since \( x = 1 \) corresponds to the point \( (1, -1) \), we disregard this and take \( x = -\frac{13}{24} \).

### Step 6: Find the corresponding \( y \)-coordinate.
Substitute \( x = -\frac{13}{24} \) into the equation of the normal line to find \( y \):

\[
y = -\frac{1}{6} \left(-\frac{13}{24} \right) - \frac{5}{6}
\]
\[
y = \frac{13}{144} - \frac{120}{144}
\]
\[
y = \frac{-107}{144}
\]

Thus, the second point of intersection is \( \left( -\frac{13}{24}, \frac{-107}{144} \right) \).

Would you like more details on any of these steps?

Here are some related questions:
1. How do we find the equation of a line tangent to a curve?
2. What is the importance of the negative reciprocal for perpendicular lines?
3. Could we solve this problem geometrically instead of algebraically?
4. What if the curve were cubic instead of quadratic? Would the method change?
5. How can we use calculus to confirm that the line is normal to the curve at a specific point?

**Tip:** Always remember that the slope of a normal line is the negative reciprocal of the slope of the tangent line!

At what point does the line normal to the graph of y = 4x^2 - (3 + 2x) at (1, -1) intersect the parabola a second time?

This problem involves finding the second intersection point of a normal line to the parabola y = 4x^2 - (3 + 2x) at (1, -1). It covers key calculus and algebra concepts like derivatives, tangent lines, normal lines, and solving quadratic equations.

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