# maths graphs helpWatch

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#1
Heyyyyyy

So, I don't get this:

g(x) = x^3 + kx^2 - 15x + 5
g'(x) = 3x^2 + 2kx - 15
m = 21
k = 6

Find the x-coordinates of the stationary points of the graph of y = g(x).

Last edited by mikaelalrc; 1 week ago
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1 week ago
#2
So a stationary point is where do/dx = 0. In this case you’re setting y=x^3+6x^2-15x+5
So you want to set your g’(x) = 0 and solve from there to find your x coordinates, and then sub into g(x) for your y coordinates.

Hope that makes sense?
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#3
hey,
yea thx soooo much, I think I get it!
Only problem is, I got the same answer as before (x=1), which is apparently wrong...

(Original post by JackMoseley)
So a stationary point is where do/dx = 0. In this case you’re setting y=x^3+6x^2-15x+5
So you want to set your g’(x) = 0 and solve from there to find your x coordinates, and then sub into g(x) for your y coordinates.

Hope that makes sense?
0
1 week ago
#4
(Original post by mikaelalrc)
hey,
yea thx soooo much, I think I get it!
Only problem is, I got the same answer as before (x=1), which is apparently wrong...
So there’s definitely for the graph y=x^3 + 6x^2 - 15x + 5
Is a minimum at x = 1, and a maximum at x = -5

How did you work out k?
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#5
Hey again!

Haha im so confused, how did you get the maximum and minimums???? And btw what did you say = 0 ?

I worked out k in a previous question, and I got it right, when they gave me:
the tangent to the graph of y = g(x) at x=2 is parallel to the line y= 21x + 7. Show that k=6.
I substituted in x=2 to my g'(x) and put in that m=21 and solved for k.

(Original post by JackMoseley)
So there’s definitely for the graph y=x^3 + 6x^2 - 15x + 5
Is a minimum at x = 1, and a maximum at x = -5

How did you work out k?
0
1 week ago
#6
is this a GCSE or Alevel question?

(Original post by mikaelalrc)
Heyyyyyy

So, I don't get this:

g(x) = x^3 + kx^2 - 15x + 5
g'(x) = 3x^2 + 2kx - 15
m = 21
k = 6

Find the x-coordinates of the stationary points of the graph of y = g(x).

0
#7
by the way, I'M STUCK ON THE NEXT ONE TOO, yay!
any help again please?? (it's question 2g and the previous is 2d so same info.

and g'(-1) obviously = -24 and g is decreasing at x= -24 as it is negative.

So, find the y-coordinate of the local minimum.
Any help pleeeease????
0
1 week ago
#8
(Original post by mikaelalrc)
Hey again!

Haha im so confused, how did you get the maximum and minimums???? And btw what did you say = 0 ?

I worked out k in a previous question, and I got it right, when they gave me:
the tangent to the graph of y = g(x) at x=2 is parallel to the line y= 21x + 7. Show that k=6.
I substituted in x=2 to my g'(x) and put in that m=21 and solved for k.

Hi!

So to find the maximum and minimum you should find the second derivative, if d^2y/dx^2 < 0 it is a maximum and if it is > 0 it is a minimum. But instead of this I actually cheated for time and put it into both Desmos.com and Wolfram Alpha.

To find the stationary points though you would set g’(x) = 0

Hmmm, not entirely sure why x=1 is incorrect then as it definitely is a stationary point of this graph!

1
#9
IB, so kinda like A-levels (but it's maths studies so like GCSE higher or extension)
(Original post by help_me_learn)
is this a GCSE or Alevel question?
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1 week ago
#10
(Original post by help_me_learn)
is this a GCSE or Alevel question?
This is an A/AS level question.
0
1 week ago
#11
(Original post by mikaelalrc)
by the way, I'M STUCK ON THE NEXT ONE TOO, yay!
any help again please?? (it's question 2g and the previous is 2d so same info.

and g'(-1) obviously = -24 and g is decreasing at x= -24 as it is negative.

So, find the y-coordinate of the local minimum.
Any help pleeeease????
Hi!

This is how I would solve this problem. Both proving that at x = 1 is a minimum (there is other ways to show this if you would like me to show you in a different way).

To find the y coordinate in this question you just need to sub the x coordinate of the minimum into the original g(x) equation.
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1 week ago
#12
thank u
(Original post by mikaelalrc)
IB, so kinda like A-levels (but it's maths studies so like GCSE higher or extension)
0
1 week ago
#13
thank u
(Original post by JackMoseley)
This is an A/AS level question.
0
#14
Thank you so much, I'll try this out, just give me a minute .

I don't have the correct answers. We had a maths exam last week and my teacher marked them (but of course only right or wrong, no explanations or correct answers). Our homework for this week is to correct our own answers, but I'm still stuck since, as I didn't get it last week in the exam, why would I get it now?
So anyway, I don't have the correct answers, sorry, I just know that x=1 is wrong, unless of course my teacher somehow was mistaken whilst marking it...
(Original post by JackMoseley)
Hi!

So to find the maximum and minimum you should find the second derivative, if d^2y/dx^2 < 0 it is a maximum and if it is > 0 it is a minimum. But instead of this I actually cheated for time and put it into both Desmos.com and Wolfram Alpha.

To find the stationary points though you would set g’(x) = 0

Hmmm, not entirely sure why x=1 is incorrect then as it definitely is a stationary point of this graph!

0
1 week ago
#15
(Original post by mikaelalrc)
Thank you so much, I'll try this out, just give me a minute .

I don't have the correct answers. We had a maths exam last week and my teacher marked them (but of course only right or wrong, no explanations or correct answers). Our homework for this week is to correct our own answers, but I'm still stuck since, as I didn't get it last week in the exam, why would I get it now?
So anyway, I don't have the correct answers, sorry, I just know that x=1 is wrong, unless of course my teacher somehow was mistaken whilst marking it...
Oh! No worries. I haven’t got a reason as to why x=1 is incorrect! As shown by Desmos etc it definitely is a stationary point along with -5!

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#16
(Original post by JackMoseley)
Hi!

This is how I would solve this problem. Both proving that at x = 1 is a minimum (there is other ways to show this if you would like me to show you in a different way).

To find the y coordinate in this question you just need to sub the x coordinate of the minimum into the original g(x) equation.
OMG I think I get it now! Thank you soooo much, soooo helpful!!!!!

(Woah, I dont think Ive ever heard thunder as loud as what I can hear outside my window rn hhahaha kinda cool actually)
1
1 week ago
#17
(Original post by mikaelalrc)
OMG I think I get it now! Thank you soooo much, soooo helpful!!!!!

(Woah, I dont think Ive ever heard thunder as loud as what I can hear outside my window rn hhahaha kinda cool actually)
Glad you get it. It’s strange as I don’t mind doing them like now, come a couple days time with my a level exams I bet I struggle 😂

(I actually love thunder, shame it’s just pouring with rain here 😅)
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#18
without using Desmos, how do you know that -5 is a stationary point?

(Original post by JackMoseley)
Oh! No worries. I haven’t got a reason as to why x=1 is incorrect! As shown by Desmos etc it definitely is a stationary point along with -5!

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#19
Oh I'm sure you'll do fine, good luck though!
(Original post by JackMoseley)
Glad you get it. It’s strange as I don’t mind doing them like now, come a couple days time with my a level exams I bet I struggle 😂

(I actually love thunder, shame it’s just pouring with rain here 😅)
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1 week ago
#20
(Original post by mikaelalrc)
without using Desmos, how do you know that -5 is a stationary point?
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.
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