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

1 Answer
e52fa787
Sep 1, 2017

Tangent: #y=2x-11#

Normal: #y=-1/2x-6#

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-2#

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

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

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

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

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

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+c#

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

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

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

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