The equation of the tangent to the curve $y = x^{3} - 6 x + 5$ at $(2, 1)$ is

6 x - y - 11 = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

6 x - y - 13 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6 x + y + 11 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6 x - y + 11 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $6 x - y - 11 = 0$
The equation of the curve
$y = x^{3} - 6 x + 5$
$\Rightarrow \frac{d y}{d x} = 3 x^{2} - 6$
$\Rightarrow {(\frac{d y}{d x})}_{2, 1} = 6$
Now, equation of the tangent at $(2, 1)$ is
$(y - 1) = 6 (x - 2)$
$\Rightarrow 6 x - y - 11 = 0$

Suggest Corrections

Similar questions

x^{2} + y^{2} + 6 x + 8 y - 11 = 0

x^{2} + y^{2} - 6 x - 14 y + 48 = 0

x^{2} + y^{2} - 6 x = 0

α, β, γ

x^{3} + 3 x^{2} + 2 = 0

\frac{α}{β + γ}, \frac{β}{γ + α}, \frac{γ}{α + β}

\sqrt{10}

2 x + y = 5

x^{2} + y^{2} - 6 x + 2 y = 0

2 x + y = 5

x^{2} + y^{2} - 6 x + 2 y = 0.

(x + y) + λ (2 x - y + 1) = 0

(1, - 3)

Join BYJU'S Learning Program