f’(x) =2x^2+kx-12 There is a stationary point at (3,-10). What is the other stationay point.
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.
`x^2-3x+2x - 6=0`
`x=3` and `x=-2`
So the other stationary point is at x=-2