Question

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

Answer 1

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

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