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Curves and turning points

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
Reply 1
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?
Reply 2
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?
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?
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.
Reply 5
60891EE6-09D1-4C0F-95B1-03630A630FE1.jpeg60891EE6-09D1-4C0F-95B1-03630A630FE1.jpeg
Reply 6
Original post by A0W0N
60891EE6-09D1-4C0F-95B1-03630A630FE1.jpeg

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?
Reply 7
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
Reply 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 redid it and got 2+2x-12x^2
I have no idea how i got that answer i apologize
Reply 9
Original post by A0W0N
i redid it and got 2+2x-12x^2
I have no idea how i got that answer i apologize

image.jpg
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
(edited 4 years ago)
Reply 10
I figured it out thanks everyone for the help!
(edited 4 years ago)

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