What degree is this polynomial? Also are there any other roots?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.
lol, I don't know what you're trying to say...What degree is this polynomial? Also are there any other roots?
The possibilities are virtually endless, but here's a typical example of what it might look like: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 he is asking is for the power of the leading term of the polynomial, e.g. x^2, x^3...lol, I don't know what you're trying to say...
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?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
... 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.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?
Here it is, it's question 5 on HSC Mathematics 2003.... 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.
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 freeHere 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?
Where did I say (0,0) was a minimum?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
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)Where did I say (0,0) was a minimum?
If (3, -27) was a maximum, how would that look like? Or is that not possible?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)
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.If (3, -27) was a maximum, how would that look like? Or is that not possible?
Okay thanks for your help, even though PhysicsMaths was the one who provided the best answer including a picture.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.