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Find the equation of the normal to the curve y = \(3x^2 + 2\...

Find the equation of the normal to the curve y = 3x2+2 at point (1, 5)

A.

6y - x - 29 = 0

B.

6y + x - 31 = 0

C.

y - 6x - 1 = 0

D.

y - 6x + 1 = 0

Correct answer is B

y = 3x2+2

y1=dydx=6x

Evaluating this derivative at x = 1 (since the point of interest is (1, 5)) gives us the slope of the tangent line at that point:
mtangent=y1(1)=6(1)=6
Slope of the Normal Line mnormal=1mtangent

mnormal=16

y−y_1​= ^mnormal⋅(x−x_1)

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

=y-5=-\frac{1}{6}x+\frac{1}{6}

=y=-\frac{1}{6}x+\frac{1}{6}+5

Multiply through by 6

=6y=-x+1+30

∴6y+ x - 31=0