At what point on the curve y = x - 3x^2 + 2 is the tangent parallel to the x-axis.
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Tushar Chandra
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The slope of a tangent drawn at any point (a, f(a)) to a curve described by y = f(x) is given by the value of f'(a).
For the curve y = x - 3x^2 + 2, the derivative y' = 1 - 6x. If a line is parallel to the x-axis, the slope of the line is 0.
To determine the point required in the problem solve y' = 0 for x.
1 - 6x = 0
=> x = 1/6
y = 1/6 - 3*(1/36) + 2 = 25/12
At the point `(1/6, 25/12)` the tangent to the curve y = x - 3x^2 + 2 is parallel to the x-axis.
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lochana2500 | Student
`y= x - 3x^(2) +2 `
`dy/dx` `= 1-6x`
`dy/dx` `=0` , when curve parallel to the x-axis.
`0=1-6x`
`x=1/6`
`when x=1/6, `
` ``y = 1/6 - 3*(1/6)^(2) +2` = `(25)/(12)`
point is `(1/6 , 25/12)`
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