Question

Find the equation of the normal to the curve y = x^3 - x^2 at point (1,1)

Answer 1

Answer on Question #83006 – Math – Analytic Geometry

Question

Find the equation of the normal to the curve $y = x^3 - x^2$ at point (1,0).

Solution

$$y = x^3 - x^2$$ $$y' = 3x^2 - 2x.$$

Slope of the tangent line:

$$m = y'(1) = 3 - 2 = 1.$$

Slope of the normal line:

$$n = -\frac{1}{m} = -1.$$

Equation of the normal line:

$$y - 0 = -1 * (x - 1), \text{ that is, } \quad y = -x + 1.$$

Answer: $y = -x + 1$

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