y = x^4 - 8x^2 + 3
dy / dx = 4x^3 - 16x
When the x coordinate is 1, the related y coordinate is (1)^4 - 8(1)^2 + 3 = -4
The slope at x = 1 = 4(1)^3 - 16(1) = -12
A normal line at this point will have a negative reciprocal slope of 1 / 12
And using the point (1, -4)...the equation of the normal line is
y =(1/12)(x -1) - 4
y = (1/12)x -1/12 - 4
y = (1/12)x -1/12 - 48/12
y = (1/12)x - 49/12 multiply through by 12
12y = 1x - 49 rearrange as
1x - 12y - 49 = 0
Here's the graph : https://www.desmos.com/calculator/gwc3nmrm5i
