Find the equation of normal lines to the curve y = x^4 + 2x^2 at (1,3).

To find the equation of the normal line to the curve y = x^4 + 2x^2 at the point (1,3), we need to follow these steps:

Step 1: Find the derivative of the given curve. The derivative of y = x^4 + 2x^2 can be found using the power rule of differentiation. Taking the derivative of each term, we get: dy/dx = 4x^3 + 4x

Step 2: Find the slope of the tangent line. To find the slope of the tangent line at the point (1,3), we substitute x = 1 into the derivative: dy/dx = 4(1)^3 + 4(1) = 4 + 4 = 8

Step 3: Find the slope of the normal line. The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is -1/8.

Step 4: Use the point-slope form to find the equation of the normal line. Using the point-slope form of a linear equation, we have: y - y1 = m(x - x1) where (x1, y1) is the given point (1,3) and m is the slope of the normal line (-1/8). Substituting the values, we get: y - 3 = (-1/8)(x - 1)

Step 5: Simplify the equation. To simplify the equation, we can distribute -1/8 to the terms inside the parentheses: y - 3 = (-1/8)x + 1/8

Finally, we can rearrange the equation to get it in the standard form: 8y - 24 = -x + 1

So, the equation of the normal line to the curve y = x^4 + 2x^2 at the point (1,3) is 8y + x = 25.