Question

Find the equation to the tangent at (0, 0) on the curve y = 4x² – 2x³

Answer 1

 y = 4x² – 2x³

`dy/dx = 8x - 6x^2`

Let P(x₁, y₁) be the point of contact.

∴ Slope of the tangents = `8x_1 - 6x_1^2`

Tangent equation: y – y₁ = `(8x_1 - 6x_1^2)(x - x_1)`

∵ The tangent passes through the origin

`y_1 = (8x_1 - 6x_1^2)x_1`

Now, `y_1 = 4x_1^2 - 2x_1^3`

= `8x_1^2 - 6x_1^3`

`\implies` 4 – 2x₁ = 8 – 6x₁

`\implies` 4x₁ = 4

`\implies` x₁ = 1

`\implies` y₁ = 2

The point on the curve is (1, 2)