# At what point on the curve y = x - 3x^2 + 2 is the tangent parallel to the x-axis.

*print*Print*list*Cite

### 2 Answers

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.

`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)`