Question

(1 point) find an equation of the tangent line to the curve y=2−2x−3x2 at (1,−3).

Answer 1

y = -8x + 5

To obtain the equation we require a point on it and it's slope.

the slope m = dy/dx at x = 1

dy/dx = -2-6x

x = 1 then dy/dx = -2 -6 = -8

equation of line in form y - b = m(x-a) with m = -8 and (a,b)=(1,-3)

y + 3 = -8(x-1)

y + 3 =-8x + 8

⇒ y = -8x +5 is the equation of the tangent