Question

Find the equation of tangent to the curve y = 3x² − x + 1 at P(1, 3).

Answer 1

Equation of the curve is y = 3x² − x + 1

Differentiating w.r.t. x, we get

`dy/dx = 6x - 1`

∴ Slope of tangent at P(1, 3) is

`(dy/dx)_((1","3))` = 6(1) − 1 = 5

∴ the equation of tangent at P(1, 3) is

y − 3 = 5(x − 1)

∴ y − 3 = 5x − 5

∴ 5x − y − 2 = 0