anonymous
  • anonymous
Confirmation and help: A curve has the equation: y= x^3 +3x^2 -9x + k, where k is a constant (i) write down an expression for dy/dx My solution: dy/dx= 3x^2 + 6x - 9 Is it correct? (ii) Find the x-coordinates of the two stationary points on the curve. My solution x=1 or x=-3 Is it correct? (iii) Hence find the two values for K for which the curve has a stationary point on the x-axis. Help please.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Yeah that's right. For (iii), remember that k disappears when you take the derivative. So the stationary points are the same that you found in (ii). So, now find what y = at those two values of x. Then, set k so that y = zero at those stationary points. I found that when x = 1, k had to be 5, and when x = -3 k has to be -27. I just rushed through, so might not be right... but I think that's the general idea... It does seem to say that you'll need two different values for k.
anonymous
  • anonymous
I'm confused... Can you show it in steps?
anonymous
  • anonymous
Well you've got (i) and (ii) right, you don't need help with that, right? For the stationary point, I think that is when the derivative equals zero or something? I've forgotten that terminology, but that's what you've said it is, and I think that's right. So, the stationary points are the two found in (ii). The question asks you to choose a value for k so that the stationary point is on the x axis. That means that y = 0. So, calculate the value of y when x equals the stationary points, which is x = 1 and x = -3. You'll get some number plus k. Set k so that the sum of the two is equal to zero. This means that the stationary point is on the x axis. Maybe that just sounds more confusing...

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anonymous
  • anonymous
Yes... Can you maybe show the steps out...?
anonymous
  • anonymous
If you read what I wrote then it should be clear.
anonymous
  • anonymous
I did, but my brain is more visual... I can't comprehend.... If you write it out in steps, bit by bit, it might help
anonymous
  • anonymous
|dw:1332847040377:dw| The two x's are the stationary points of the function y. You want to calculate what function is equal to at those points, and set k equal to the negative of that value, so that the graph is moved up or down so that the point x occurs on the x axis.
anonymous
  • anonymous
Very good :) Thanks! lol, when I meant visual, however, I meant sum visual... eg. \[{dy \over dx} = 3x^2 + 6x -9 +k\]\[0= 2x^2+6x-9+k....?\]
anonymous
  • anonymous
no, dy/dx = 3x^2 + 6x - 9. k is a constant, so it disappears when you take the derivative.
anonymous
  • anonymous
Yes... I was giving an example.... So, how do I do this?
anonymous
  • anonymous
Therefore dy/dx = 0 when x = 1 or -3, same as before. So now y has stationary points x = 1 and x = -3. What is the value of y when x = 1 or x = -3? y = (1)^3 + 3(1)^2 + .... + k = -5 + k, The question asks to set k so that y = 0 at the critical points... so above you want -5 + k = 0... k = .... 5. Do the same thing for when x = -3.
anonymous
  • anonymous
But.. 3(1)^2 + 3(1) -9(1) + k= 0+k.? maybe I'm just not understanding>>?
anonymous
  • anonymous
@Hero @scarydoor
anonymous
  • anonymous
(iii) Hence find the two values for K for which the curve has a stationary point on the x-axis.

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