6y - x - 29 = 0
6y + x - 31 = 0
y - 6x - 1 = 0
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
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