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mylen [45]
2 years ago
11

Find the equation of the normal to the curve y=x^2-2x^2-4x+1 at thepoint (-1,2)​

Mathematics
1 answer:
balandron [24]2 years ago
5 0

Answer:

y=-\frac{1}{3}x+\frac{5}{3}

Step-by-step explanation:

Assuming that the first exponent in the formula for the curve should be 3, not 2...

y=x^3-2x^2-4x+1

The derivative is

y'=3x^2-4x+4

The slope of the tangent line at the point (-1, 2) is the value of the derivative at x = -1.

y'|_{x=-1} = 3(-1)^2-4(-1)+4=3-4+4=3

The slope of the normal line is the opposite reciprocal of the slope of the tangent line.

m_{\text{normal} = -\frac{1}{3}

Using the Point-Slope form of a linear equation, the normal line is

y-2=-\frac{1}{3}(x-(-1))\\y-2=-\frac{1}{3}(x+1)\\y=-\frac{1}{3}x-\frac{1}{3}+2\\y=-\frac{1}{3}x+\frac{5}{3}

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