# Curves and turning pointsWatch

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Thread starter 3 weeks ago
#1
Equation of curve: y=2x+x^2 -4^3
I found the turning points and they're (-1,-4) and (-3,0)
But whem i put thw equation into a graphing calculator there are different points. I differentiated the equation to get 3x^2 +12x+9.
I appreciate any help, Thanks
0
3 weeks ago
#2
(Original post by A0W0N)
Equation of curve: y=2x+x^2 -4^3
I found the turning points and they're (-1,-4) and (-3,0)
But whem i put thw equation into a graphing calculator there are different points. I differentiated the equation to get 3x^2 +12x+9.
I appreciate any help, Thanks
That differential doesn't match your curve equation. Can you check the question and let us know what it says?
0
Thread starter 3 weeks ago
#3
(Original post by Pangol)
That differential doesn't match your curve equation. Can you check the question and let us know what it says?
Yes thats the correct equation have i differentiated it wrong?
0
3 weeks ago
#4
(Original post by A0W0N)
Yes thats the correct equation have i differentiated it wrong?
As Pangol said, the curve doesn't match the deriviative, but more than that the initial equation looks weird with the -4^3 at the end. Can you upload an image of the original question so we can be certain?
0
3 weeks ago
#5
(Original post by A0W0N)
Equation of curve: y=2x+x^2 -4^3
I found the turning points and they're (-1,-4) and (-3,0)
But whem i put thw equation into a graphing calculator there are different points. I differentiated the equation to get 3x^2 +12x+9.
I appreciate any help, Thanks
At any rate, well done for having the sense to check your answer with a graphics calculator. In the exam this would have helped you spot your mistake, maybe correct it, and get a few more marks. Far too many ignorant teachers tell students that GDCs aren't worth having. You've just provided one (of very many) example of how important it is to have one, and know how to use it.
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Thread starter 3 weeks ago
#6
0
3 weeks ago
#7
(Original post by A0W0N)
That really helps. Can you explain how you get your dy/dx? You say you get 3x^2 +12x+9 - how do you get this?
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Thread starter 3 weeks ago
#8
(Original post by Pangol)
That really helps. Can you explain how you get your dy/dx? You say you get 3x^2 +12x+9 - how do you get this?
I multiplyed it by the powers then multiplyed it by (-1) so i could try and factorise it
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Thread starter 3 weeks ago
#9
(Original post by Pangol)
That really helps. Can you explain how you get your dy/dx? You say you get 3x^2 +12x+9 - how do you get this?
i redid it and got 2+2x-12x^2
I have no idea how i got that answer i apologize
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Thread starter 3 weeks ago
#10
(Original post by A0W0N)
i redid it and got 2+2x-12x^2
I have no idea how i got that answer i apologize

Edit: I now have the correct turning points i just ont know how to find the the nature of them and i can't differenciate twice
Last edited by A0W0N; 3 weeks ago
0
Thread starter 3 weeks ago
#11
I figured it out thanks everyone for the help!
Last edited by A0W0N; 3 weeks ago
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