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# Find the slope int. equation of the normal line to the graph f(x)=x^3-x^2 at the point where x=1?

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- Marley KLv 74 weeks agoFavourite answer
first find the slope of the tangent line by taking the derivative:

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

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

f'(1) = 3 - 2 = 1 <<<====slope of tangent line where x=1

the normal line is perpendicular to the tangent, so the slope of the normal line is -1.

find the y-coordinate:

f(1) = 1 - 1 = 0

Now you have the slope, m, and point (1,0) and can use that to find the equation:

y = mx + b

0 = (-1)(1) + b

b = 1

equation in slope-intercept form:

y = -1x + 1

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