Question #627ac

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Answer 1 · 2017-11-15T20:55:34

$m = - \frac{1}{2}$ $m=-1/2$

we take

$x^{3} y^{3} = 8$ $x^3y^3=8$

taking out cube root

$x y = 2$ $xy=2$

isolate the variable y

$y = \frac{2}{x}$ $y=2/x$

deriving to find the slope of the line tangent to the curve

$y'=-2/x^2$

on the point $(2,1) -> x=2$

substitute in the derivative

$y'=-2/(2^2)$

$y'=-1/2$

the tangent line will be given by the point-slope equation

$y-y_o=m(x-x_o)$

$y-1=-1/2(x-2)$

operating

$x+2y-4=0$