Asked by Millie
find the horizontal tangents of the curve and show how.
y=x^3-2x^2+x+1
y=x^3-2x^2+x+1
Answers
Answered by
Steve
y = x^3 - 2x^2 + x + 1
Horizontal tangents have slope = 0
y' = 3x^2 - 4x + 1 = (3x-1)(x-1)
so, at x = 1/3 or 1 the slope is zero.
So, the lines are
y = 13/9
y = 1
Horizontal tangents have slope = 0
y' = 3x^2 - 4x + 1 = (3x-1)(x-1)
so, at x = 1/3 or 1 the slope is zero.
So, the lines are
y = 13/9
y = 1
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