# f’(x) =2x^2+kx-12 There is a stationary point at (3,-10). What is the other stationay point.

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It is given that `f'(x) =2x^2+kx-12`

When we have a stationary point `f'(x) = 0`

`f'(x) = 0`

`2x^2+kx-12 = 0`

It is given that at `P(3,-10)` we have a stationary point.

`[f'(x)]_(x=3) =0`

`2(3)^2+3k-12=0`

`18+3k-12=0`

`k= -2`

`2x^2-2x-12=0`

`x^2-x-6=0`

`x^2-3x+2x - 6=0`

`x(x-3)+2(x-3)=0`

`(x-3)(x+2)=0`

`x=3` and `x=-2`

*So the other stationary point is at x=-2*

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