# calculus

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determine the point of intersection of the tangents at the points of inflection to the curve f(x)= x^4 – 24x^2 – 2

• calculus -

y' = 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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