SOLUTION: The curve whose equation is y= (2x2+3s-9)(x-k), where k is a constant, has a turning point where x=(-1) 1. Calculate the value of k and find the value of x at the other turning

Question 250882: The curve whose equation is y= (2x2+3s-9)(x-k), where k
is a constant, has a turning point where x=(-1)
1. Calculate the value of k and find the value of x at the other turning point of the curve
Answer by Greenfinch(383) (Show Source): You can put this solution on YOUR website!
It doesn't seem to make entire sense to me like this,
could it be (2x^2 + 3x -9)(x - k)?
Then expanding gives (2x^3 + 3x^2 -9x) - (2kx^2 + 3kx -9k)
In powers of x this is (2)x^3 + (3-2k)x^2 -(9+3k)x -9k
The turning point requires dy/dx = 6x^2 + 2(3-2k)x - (9+3k)=0
This is a turning point at x = -1, so substituting
6 -(6 - 4k) - (9 + 3k) = -9 -k = 0 or k = 9
Putting this in dy/dx gives 6x^2 +30x -36 =0
Divide by 6 x^2 + 5x - 6 =0
which factorises to (x + 1)(x - 6) = 0
x = -1 or 6 [ we were given the -1 so this fits]

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