# maths graphs helpWatch

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#21
ok... thank you!
(Original post by JackMoseley)
You can find all stationary points by just solving dy/dx = 0

As your g’(x) is a quadratic, you’ll get two roots, one being 1 and the other being -5 in this example.

If there is only 1 stationary point, the gradient function would not be a quadratic. Sometimes a simple sketch of a graph if you can can be useful.
0
2 weeks ago
#22
(Original post by mikaelalrc)
ok... thank you!
Hope that makes sense still. I can be known to not make sense 😂
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#23
ahahhah dont worry, i suck at explaining things most of the time!

I think it makes sense, Im just super tired now though ahhah maybe what you just explained will click for me tomorrow!

THANK YOU SO MUCH FOR YOUR HELP, muuuuuch appreciated!!!
goodnight!!!!!!
(Original post by JackMoseley)
Hope that makes sense still. I can be known to not make sense 😂
0
2 weeks ago
#24
(Original post by mikaelalrc)
ahahhah dont worry, i suck at explaining things most of the time!

I think it makes sense, Im just super tired now though ahhah maybe what you just explained will click for me tomorrow!

THANK YOU SO MUCH FOR YOUR HELP, muuuuuch appreciated!!!
goodnight!!!!!!
Haha that would make sense!

No worries, if you ever need any other help I don’t mind if you pm me, I’m more than happy to help!

Goodnight 👍
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#25
you definitely make sense! by the way, will you be around here on TSR tomorrow or sometime before next Monday? I'll have loads more questions then...!
(Original post by JackMoseley)
Hope that makes sense still. I can be known to not make sense 😂
0
2 weeks ago
#26
(Original post by mikaelalrc)
you definitely make sense! by the way, will you be around here on TSR tomorrow or sometime before next Monday? I'll have loads more questions then...!
Of course, I’m always on and off every day basically, so will be able to answer anything at any point basically
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