Find the equation of the normal to y = 2x3 – x2 + 3 at (1, 4).

1.	Find the equation of the normal to y = 2x3 – x2 + 3 at (1, 4).
Answer» Given as y = 2x³- x² + 3 Differentiate with respect to x, to get the slope of tangent m = dy/dx = 6x² - 2x m = 4 at (1,4) The normal is perpendicular to tangent therefore, m₁m₂ = -1 m(normal) = -1/4 The equation of normal is given by y - y₁ = m(normal)(x - x₁) y - 4 = (-1/4)(x - 1) x + 4y = 17

Answer»

Given as y = 2x³- x² + 3

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

m = dy/dx = 6x² - 2x

m = 4 at (1,4)

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

m(normal) = -1/4

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

y - 4 = (-1/4)(x - 1)

x + 4y = 17

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