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Find Equation of a Normal Line?
Trying to solve the equation below:
y = (2x - 1)^3
slope of -1/24, x > 0
But I don't even come close to the answer of y - 8 = -1/24 (x - 3/2) or 2x + 48y - 387 = 0
1 Answer
- ?Lv 78 years agoFavorite Answer
The gradient of the normal is -1/(gradient of the tangent)
(d/dx)y = 6(2x - 1)^2 = 24 where x = -1/2, and x= 3/2 , but we already know we are only interested in x>0, so x=3/2 it is.
From there y = (2(3/2) - 1)^3 = 8 and we already know that m = -1/24 so (y-8) =(-1/24)(x-3/2) [point slope form of the equation of a line], and hence y = -x/24 + 129/16, which is also 2x + 48y - 387 = 0.