Find the equation of the tangent and the normal to the curves at the i

1.	Find the equation of the tangent and the normal to the curves at the indicated points: y = x4 – 6x3 + 13x2 – 10x + 5 at x = 1 and y = 3
Answer» Given as y = x⁴ – 6x³ + 13x² – 10x + 5 at x = 1 and y = 3 Differentiate with respect to x, to get the sloe of tangent dy/dx = 4x³ - 18x² + 26x - 10 m(tangent) at (x = 1) = 2 The normal is perpendicular to tangent therefore, m₁m₂ = -1 m(normal) at (x = 1) = -1/2 The equation of tangent is given by y - y₁ = m(tangent)(x - x₁) y - 3 = 2(x - 1) y = 2x + 1 The equation of normal is given by y - y₁ = m(normal)(x - x₁) y - 3 = (-1/2)(x - 1) 2y = 7 - x

Find the equation of the tangent and the normal to the curves at the indicated points: y = x4 – 6x3 + 13x2 – 10x + 5 at x = 1 and y = 3

Answer»

Given as y = x⁴ – 6x³ + 13x² – 10x + 5 at x = 1 and y = 3

Differentiate with respect to x, to get the sloe of tangent

dy/dx = 4x³ - 18x² + 26x - 10

m(tangent) at (x = 1) = 2

The normal is perpendicular to tangent therefore, m₁m₂ = -1

m(normal) at (x = 1) = -1/2

The equation of tangent is given by y - y₁ = m(tangent)(x - x₁)

y - 3 = 2(x - 1)

y = 2x + 1

The equation of normal is given by y - y₁ = m(normal)(x - x₁)

y - 3 = (-1/2)(x - 1)

2y = 7 - x