Find the equations of the tangent and normal to the curve y=x^3-2x^2-2x-3 at x=2?

1 Answer
Sep 1, 2017

Tangent: y=2x-11y=2x11

Normal: y=-1/2x-6y=12x6

Explanation:

A tangent line would have a gradient equal to that of the curve at x=2, while a normal line would have a gradient which is the negative reciprocal of that of the curve at x=2

dy/dx=3x^2-4x-2dydx=3x24x2

When x=2, dy/dx=3(2)^2-4(2)-2=2dydx=3(2)24(2)2=2

Hence, tangent at x=2: y=(dy/dx)x+c=2x+cy=(dydx)x+c=2x+c

Solve for c by setting x=2, y=(2)^3-2(2)^2-2(2)-3=-7x=2,y=(2)32(2)22(2)3=7

Thus, c=-7-2(2)=-11c=72(2)=11

The equation of the tangent line is thus y=2x-11y=2x11

graph{(y-x^3+2x^2+2x+3)(y-2x+11)=0 [1, 3, -7.5, -6.5]}

Similarly, for the normal line, y=-1/2x+cy=12x+c

c=-7-(-1/2(2))=-6c=7(12(2))=6

Thus the equation of the normal line is y=-1/2x-6y=12x6

graph{(y-x^3+2x^2+2x+3)(y+1/2x+6)=0 [1, 3, -7.5, -6.5]}