Question

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

Answer 1

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]}