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November 26, 2014

November 26, 2014

Posted by **samantha** on Thursday, March 28, 2013 at 5:00pm.

- calculus -
**Reiny**, Thursday, March 28, 2013 at 9:14pmy' = 4x^3 - 48x

y '' = 12x^2 - 48

at points of inflection, y'' = 0

12x^2 = 48

x^2 = 4

x = ±2

when x=2 , f(2) = 16 - 96 - 2 = -82 , slope = 32 - 96 = -64

when x = -2, f(-2) = -82 , slope = -32 + 96 = 64

1st tangent: slope = -64, point is (2,-82)

-82 = -64(2) + b

b=46

first tangent equation: y = -64x + 46

2nd tangent: slope = 64 , point is (-2,-82)

-82 = 64(-2) + b

b = 46

second tangent is y = 64x + 46

intersection of y = 64x + 46 and y = -64x + 46

64x + 46 = -64x + 46

128x = 0

x = 0, then y = 46

the two tangents intersect at (0,46)

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