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 = 2x2 – 3x – 1 at (1, – 2)
Answer» Given as y = 2x² – 3x – 1 at (1, – 2) Differentiate the given curve, to get the slope of the tangent dy/dx = 4x - 3 m(tangent) at (1, – 2) = 1 The normal is perpendicular to tangent therefore, m₁m₂ = – 1 m(normal) at (1, – 2) = – 1 The equation of tangent is given by y – y₁ = m(tangent)(x – x₁) y + 2 = 1(x – 1) y = x – 3 The equation of normal is given by y – y₁ = m(normal)(x – x₁) y + 2 = –1(x – 1) y + x + 1 = 0

Find the equation of the tangent and the normal to the curves at the indicated points: y = 2x2 – 3x – 1 at (1, – 2)

Answer»

Given as y = 2x² – 3x – 1 at (1, – 2)

Differentiate the given curve, to get the slope of the tangent

dy/dx = 4x - 3

m(tangent) at (1, – 2) = 1

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

m(normal) at (1, – 2) = – 1

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

y + 2 = 1(x – 1)

y = x – 3

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

y + 2 = –1(x – 1)

y + x + 1 = 0