How do you find a cubic function y=ax3+bx2+cx+d whose graph has horizontal tangents at the = points (-2,6) and (2,0)?

wurpenxd

wurpenxd

Answered question

2022-09-12

How do you find a cubic function y = a x 3 + b x 2 + c x + d
whose graph has horizontal tangents at the points (-2,6) and (2,0)?

Answer & Explanation

Duncan Kaufman

Duncan Kaufman

Beginner2022-09-13Added 17 answers

f ( x ) = 3 16 x 3 - 9 4 x + 3
Explanation:
Given
f ( x ) = a x 3 + b x 2 + c x + d the condition of horizontal tangency at points { x 1 , y 1 } , { x 2 , y 2 } is
d f d x f ( x = x 1 ) = 3 a x 1 2 + 2 b x 1 + c = 0
d f d x f ( x = x 2 ) = 3 a x 2 2 + 2 b x 2 + c = 0
also we have in horizontal tangency f ( x = x 1 ) = a x 1 3 + b x 1 2 + c x 1 + d = y 1
f ( x = x 2 ) = a x 2 3 + b x 2 2 + c x 2 + d = y 2
so we have the equation system ( 12 a - 4 b + c = 0 12 a + 4 b + c = 0 - 8 a + 4 b - 2 c + d = 6 8 a + 4 b + 2 c + d = 0 )
Solving for a , b , c , d we get ( a = 3 16 b = 0 c = - 9 4 d = 3 )

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