TANGENTS AND NORMALS

TANGENTS AND NORMALS


C
Asked by 2 months ago
10 million points

Find the equation of the tangents to the curve $y=x^3$ at points $(2,8)$.

TANGENTS AND
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1 Answer

C
Answered by 2 months ago
10 million points

$y=x^3$

$\therefore$ grad $y=3x^2$

When $x=2$,

grad $y=12$

Thus the gradient of the tangent at $(2,8)$ is $+12$. Its equation is

$\frac{y-8}{x-2}=12$

$\therefore y-8=12x-24$

$\therefore$ the equation of the tangent is $12x-y-16=0$

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Asked: 2 months ago

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