The equation of the tangent to the curve y = x3 - 6x + 5 at (2,1) is

Question

The equation of the tangent to the curve y = x3 - 6x + 5 at (2,1) is (A) 6x - y - 11 = 0 (B) 6x - y - 13 = 0 (C) 6x + y + 11 = 0 (D) 6x - y + 11 = 0

Tardigrade Admin · Accepted Answer

The equation of the curve y=x3-6 x+5 ⇒ (d y/d x)=3 x2-6 ⇒ ((d y/d x))(2,1)=6 Now, equation of the tangent at (2,1) is (y-1)=6(x-2) ⇒ y-1=6 x-12 ⇒ 6 x-y-11=0

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

6x - y - 11 = 0

6x - y - 13 = 0

6x + y + 11 = 0

6x - y + 11 = 0

x - 6y - 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
$(y - 1) = 6 (x - 2)$
$\Rightarrow y - 1 = 6 x - 12$
$\Rightarrow 6 x - y - 11 = 0$

Q. The equation of the tangent to the curve y=x3−6x+5 at (2,1) is