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y = x^3 - 3x^2 + 6x + 2
y' = 3x^2 - 6x + 6
so, when is y'=3?
3x^2-6x+6 = 3
3x^2-6x+3 = 0
3(x-1)^2 = 0
x = 1
since y(1) = 6, the tangent line there is y-6=3(x-1)
see the graphs at
https://www.wolframalpha.com/input/?i=plot+x%5E3+-+3x%5E2+%2B+6x+%2B+2%2C+y%3D3%28x-1%29%2B6
Find the coordinates of the point on the curve y = x
3 − 3x
2 + 6x + 2 at which
the gradient is 3
1 answer