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# Find the equation of the line tangent to the curve y=x³-6x² at its point of inflection.?

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- PhilipLv 61 month ago
y = x^3 -6x^2.

y' = 3x^2 -12x.

y'' = 6x -12.

Point of inflection occurs at P(a,b), where y''(a) = 0 and y(a) = b.

Putting y'' = 0 gives x = 2 and y(2) = 2^3 -6*2^2 = 8-24 = -16. Then P = (2,-16).

Slope of tangent line at P = y'(2) = 3*4 -12*2 = -12.

Equation of tangent line L at P is y = -12x +b.

P is on L. Therefore -16 = -12(2) +b, ie., b = 24-16 = 8 and eqn of L is y = -12x +8.

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- 1 month ago
Find when y'' = 0

y = x^3 - 6x^2

y' = 3x^2 - 12x

y'' = 6x - 12

y'' = 0

0 = 6x - 12

0 = x - 2

x = 2

Now, find the value of y' when x = 2

y' = 3 * 2^2 - 12 * 2 = 12 - 24 = -12

Now find the value of y when x = 2

y = 2^3 - 6 * 2^2 = 8 - 24 = -16

We need a line with a slope of -12 that passes through (2 , -16)

y - (-16) = -12 * (x - 2)

y + 16 = -12x + 24

y = -12x + 8

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