What is the equation of the line normal to f(x)=x^3-6x^2-2 f(x)=x3−6x2−2 at x=1x=1?

1 Answer
mason m
Dec 30, 2015

y+7=1/9(x-1)y+7=19(x−1)

Explanation:

Find the slope of the tangent line, to which the normal line is perpendicular.

f(1)=-7f(1)=−7

The line will pass through the point (1,-7)(1,−7).

f'(x)=3x^2-12x

f'(1)=-9

If the slope of the tangent line is -9, the slope of the normal line will be the opposite reciprocal of -9 since the lines are perpendicular. This means the slope is 1/9.

Write the equation in point-slope form:

y+7=1/9(x-1)

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