Asked by Amie
Write an equation of the line tangent to y=(9-x^2)^2/3 at x=1
Answers
Answered by
Reiny
dy/dx = (2/3)(9-x^2)^(-1/3) (2x)
when x = 1, y = 8^(2/3) = 4
dy/dx = (2/3)(8)^(-1/3) (2)
= (4/3)((1/2) = 2/3
equation:
y = (2/3)x + b
at( 1,4)
4 = (2/3) + b
b = 10/3
y = (2/3)x + 10/3
when x = 1, y = 8^(2/3) = 4
dy/dx = (2/3)(8)^(-1/3) (2)
= (4/3)((1/2) = 2/3
equation:
y = (2/3)x + b
at( 1,4)
4 = (2/3) + b
b = 10/3
y = (2/3)x + 10/3
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