Question

Find the equations of tangent and normal to the curve y = 2x³ − x² + 2 at point `(1/2, 2)`.

Answer 1

Given: y = 2x³ − x² + 2

Differentiate w. r. t. x

`dy/dx = d/dx (2x^3 - x^2 + 2)`

= 6x² − 2x

Slope of tangent at `(1/2, 2)` = m = `6(1/2)^2 - 2(1/2)`

∴ m = `1/2`

Slope of normal at `(1/2, 2)` = m' = −2

Equation of tangent is given by

`y − 2 = 1/2(x − 1/2)`

`2y - 4 = (2x - 1)/2`

4y − 8 = 2x − 1

2x − 4y + 7 = 0

Equation of normal is given by

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

y − 2 = −2x + 1

2x + y − 3 = 0