find the equation of the normal to the curve y=3x2-2x-1which
3个回答

find the equation of the normal to the curve y=3x2-2x-1which is perpendicular to the line y=x-3

Solution:

y = 3x^2 - 2x - 1 (1) Its derivative is dy/dx = 6x - 2

Let the equation of tangent line be y = mx + c1 = x + c1 (2)

(Tangent line is parallel to y = x - 3,so its gradient is 1)

Let 6x - 2 = 1,we obtain x = 1/2

Sub x = 1/2 into (1),we get y = 3/4 - 1 - 1 = - 5/4

Coordinates of tangent point are (1/2,-5/4)

Let the equation of normal line be y = -x + c2 (3)

Sub x = 1/2 and y = -5/4 into (3),

we have c2 = 1/2 - 5/4 = -1/4

Sub c2 = -1/4 back into (3),finally,the normal equation is

y = -x - 1/4 (Ans)

相关问题