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

Hint: The normal line is perpendicular to the tangent line. If two lines are perpendicular their slopes are negative reciprocals -- i.e. if the slope of the first line is $m$ then the slope of the second line is $-1 / m$

Step 1 ：Differentiate \(y = x - 4 + 2x^2\) with respect to \(x\) to get \(y' = 1 + 4x\)

Step 2 ：Substitute \(x = 1\) into \(y'\) to get \(y'(1) = 1 + 4(1) = 5\)

Step 3 ：The slope of the normal line is the negative reciprocal of the slope of the tangent line, so the slope of the normal line is \(-1/5\)

Step 4 ：Substitute the slope \(m = -1/5\) and the point \((1,-1)\) into the equation \(y = mx + c\) to get \(-1 = -1/5 * 1 + c\), then solve for \(c\) to get \(c = -1 + 1/5 = -4/5\)

Step 5 ：The equation of the normal line is \(y = -1/5x - 4/5\)

Step 6 ：Set \(y = x-4+2 x^{2}\) equal to \(y = -1/5x - 4/5\) and solve for \(x\) to get \(10x^2 + 6x - 16 = 0\)

Step 7 ：Solve this quadratic equation for \(x\) to get \(x = 1\) or \(x = -8/5\)

Step 8 ：Substitute \(x = -8/5\) into the equation of the parabola to get \(y = -8/5 - 4 + 2(-8/5)^2 = -32/25 - 100/25 + 128/125 = -128/125\)

Step 9 ：\(\boxed{\text{The line normal to the graph of } y=x-4+2 x^{2} \text{ at } (1,-1) \text{ intersects the parabola a second time at } \left(-\frac{8}{5}, -\frac{128}{125}\right)}\)