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maths graphs help

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).

Any help please?

(edited 4 years ago)

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Reply 1

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?

Reply 2

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...

Original post by JackMoseley

Reply 3

JackMoseley

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?

Reply 4

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.

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?

Reply 5

help_me_learn

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).

Any help please?

Reply 6

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???? :smile:

Reply 7

JackMoseley

Original post by mikaelalrc

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!

What answers does it give?

Reply 8

mikaelalrc

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?

Reply 9

JackMoseley

Original post by help_me_learn

is this a GCSE or Alevel question?

This is an A/AS level question.

Reply 10

JackMoseley

Original post by mikaelalrc

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.

358B7F7B-F70D-4463-9546-E49709118B6C.jpg.jpeg23.1KB

Reply 11

help_me_learn

thank u

Original post by mikaelalrc

IB, so kinda like A-levels (but it's maths studies so like GCSE higher or extension)

Reply 12

help_me_learn

thank u

Original post by JackMoseley

This is an A/AS level question.

Reply 13

mikaelalrc

Thank you so much, I'll try this out, just give me a minute :h:

.

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? :confused:

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

Reply 14

JackMoseley

Original post by mikaelalrc

Thank you so much, I'll try this out, just give me a minute :h:

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!

I’m happy to help you more if you need anything 🙂

Reply 15

mikaelalrc

Original post by JackMoseley

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 :eek:

hhahaha kinda cool actually)

Reply 16

JackMoseley

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 :eek:

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 😅)

Reply 17

mikaelalrc

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!

I’m happy to help you more if you need anything 🙂

Reply 18

mikaelalrc

Oh I'm sure you'll do fine, good luck though!

Original post by JackMoseley

Reply 19

JackMoseley

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.