find the equation of the tangent line and normal line to curve described by the following equation at point(-1,-1/5): y=(1/5)x^3

Answer 1

y = (1/5)x³

y' = (3/5)x²

At (-1, -1/5) the slope of the tangent is 3/5

At (-1, -1/5) the slope of the normal is -5/3

Use point-slope form to determine equation of each.

Answer 2

y = x³/5

y' = 3x²/5 <<== slope, m of tangent line

At point (-1,-1/5), m = 3/5

y - (-1/5) = (3/5)(x-(-1))

y = 3x/5 + 3/5 - 1/5

y = 3x/5 +2/5 <=== Tangent line

The slope of the normal line = -1/m

y + 1/5 = (-5/3)(x+1)

y = -5x/3 -5/3 + 1/5

y = -5x/3 - 28/15 <=== Normal Line