• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

How do you sketch this sort of graph? (1 Viewer)

BlueGas

Well-Known Member
Joined
Sep 20, 2014
Messages
2,448
Gender
Male
HSC
N/A
Let's say the graph has stationary points, (0,0) and (3,5). By the way, point (3,5) is a maximum, so how would you sketch a graph like this? I'm looking for answers that include pictures, thanks.
 

Librah

Not_the_pad
Joined
Oct 28, 2013
Messages
912
Location
Sydney Australia
Gender
Male
HSC
2014
Let's say the graph has stationary points, (0,0) and (3,5). By the way, point (3,5) is a maximum, so how would you sketch a graph like this? I'm looking for answers that include pictures, thanks.
What degree is this polynomial? Also are there any other roots?
 

PhysicsMaths

Active Member
Joined
Dec 9, 2014
Messages
179
Gender
Male
HSC
2015
Let's say the graph has stationary points, (0,0) and (3,5). By the way, point (3,5) is a maximum, so how would you sketch a graph like this? I'm looking for answers that include pictures, thanks.
The possibilities are virtually endless, but here's a typical example of what it might look like:
http://imgur.com/gvV6rez

Also note, that since (3,5) is a max sp,it is guaranteed that (0,0) is a min.
Also between those two points exist a pt of inflexion (change in concavity)

Other than that, it can be anything. Not enough information is given
 

PhysicsMaths

Active Member
Joined
Dec 9, 2014
Messages
179
Gender
Male
HSC
2015
lol, I don't know what you're trying to say...
What he is asking is for the power of the leading term of the polynomial, e.g. x^2, x^3...
Roots are where the curve hits the x-axis, so for my example, double root at (0,0) and single root at approx. x = 5

Single roots just cross the x-axis
Double roots sort of bounce back up
And triple roots are like pts of inflexion
 

BlueGas

Well-Known Member
Joined
Sep 20, 2014
Messages
2,448
Gender
Male
HSC
N/A
The possibilities are virtually endless, but here's a typical example of what it might look like:
http://imgur.com/gvV6rez

Also note, that since (3,5) is a max sp,it is guaranteed that (0,0) is a min.
Also between those two points exist a pt of inflexion (change in concavity)

Other than that, it can be anything. Not enough information is given
See that's the part that confuses me, how can (0,0) be a min? It's 0, it's neither a maximum or minimum. How come the answer for this type of question says that (0,0) is a horizontal point of inflexion?
 

Librah

Not_the_pad
Joined
Oct 28, 2013
Messages
912
Location
Sydney Australia
Gender
Male
HSC
2014
See that's the part that confuses me, how can (0,0) be a min? It's 0, it's neither a maximum or minimum. How come the answer for this type of question says that (0,0) is a horizontal point of inflexion?
... it's a double root. the polynomial lets say p(x)= x^2 multiplied by (......whatever else cause we don't know based on your info). If your saying it's a min point, how can it be a horizontal point of inflexion? By definition that's impossible.
 
Last edited:

BlueGas

Well-Known Member
Joined
Sep 20, 2014
Messages
2,448
Gender
Male
HSC
N/A
... it's a double root. the polynomial lets say p(x)= x^2 multiplied by (......whatever else cause we don't know based on your info). If your saying it's a min point, how can it be a horizontal point of inflexion? By definition that's impossible.
Here it is, it's question 5 on HSC Mathematics 2003.

(a) Consider the function f(x) = x^4 − 4x^3

(ii) Find the coordinates of the stationary points of the curve y = f(x), and
determine their nature.
(iii) Sketch the graph of the curve y = f(x), showing the stationary points.

Answer: (0,0) is a horizontal point of inflexion and (3, -27) is a minimum. How would you sketch this?
 

Librah

Not_the_pad
Joined
Oct 28, 2013
Messages
912
Location
Sydney Australia
Gender
Male
HSC
2014
Here it is, it's question 5 on HSC Mathematics 2003.

(a) Consider the function f(x) = x^4 − 4x^3

(ii) Find the coordinates of the stationary points of the curve y = f(x), and
determine their nature.
(iii) Sketch the graph of the curve y = f(x), showing the stationary points.

Answer: (0,0) is a horizontal point of inflexion and (3, -27) is a minimum. How would you sketch this?
Well why'd u say 0,0 was a min point... I don't know how to use this latex thing, so just go wolframalpha the eq into the system to see graph, if it's still for free
 

PhysicsMaths

Active Member
Joined
Dec 9, 2014
Messages
179
Gender
Male
HSC
2015
http://imgur.com/FcHMGLH
This is what it should look like

Note that since (3,-27) is a minimum stat. pt., the graph must go up after that
and btw, f(x) can be factorised to x^3(x-4)
and the x^3 represents the triple root (pt of inflexion) at (0,0)
After differentiating, (0,0) is also a stat. pt., so (0,0) is a horizontal pt of inflexion
 

BlueGas

Well-Known Member
Joined
Sep 20, 2014
Messages
2,448
Gender
Male
HSC
N/A
Well why'd u say 0,0 was a min point... I don't know how to use this latex thing, so just go wolframalpha the eq into the system to see graph, if it's still for free
Where did I say (0,0) was a minimum?
 

Librah

Not_the_pad
Joined
Oct 28, 2013
Messages
912
Location
Sydney Australia
Gender
Male
HSC
2014
Where did I say (0,0) was a minimum?
Hmm must have misread, my bad. But it should pass through 0,0 and have a cubic-like appearance and pass through x=4 after a min point (3,-27). Oh right this part "point (3,5) is a maximum" was this apart of the same question? Cause there are only 2 stationary points in your graph and one of them is at (3,-27)
 

BlueGas

Well-Known Member
Joined
Sep 20, 2014
Messages
2,448
Gender
Male
HSC
N/A
Hmm must have misread, my bad. But it should pass through 0,0 and have a cubic-like appearance and pass through x=4 after a min point (3,-27)
If (3, -27) was a maximum, how would that look like? Or is that not possible?
 

Librah

Not_the_pad
Joined
Oct 28, 2013
Messages
912
Location
Sydney Australia
Gender
Male
HSC
2014
If (3, -27) was a maximum, how would that look like? Or is that not possible?
You would have to completely change the equation. Assuming 0,0 was still your horizontal inflex point. You would essentially either have a curve with discontinuities or 2 or more stationary points. Better ask one of the maths pro's on this site though, i havn't done anything for the past 2-3 months.
 
Last edited:

BlueGas

Well-Known Member
Joined
Sep 20, 2014
Messages
2,448
Gender
Male
HSC
N/A
You would have to completely change the equation. Assuming 0,0 was still your horizontal inflex point. You would essentially either have a curve with discontinuities or 2 or more stationary points.
Okay thanks for your help, even though PhysicsMaths was the one who provided the best answer including a picture.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top