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

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

KEAMKEAM 2016Application of Derivatives

Solution:

The equation of the curve
$y=x^{3}-6 x+5$
$\Rightarrow \frac{d y}{d x}=3 x^{2}-6$
$\Rightarrow \left(\frac{d y}{d x}\right)_{(2,1)}=6$
Now, equation of the tangent at $(2,1)$ is
$(y-1)=6(x-2)$
$\Rightarrow y-1=6 x-12$
$\Rightarrow 6 x-y-11=0$