Hi friends,


I trust you are all doing well?....please would someone just explain this very simple thing for me?..The question is..:


Graph \(y=f'(x)=mx^2+nx+k\) goes through the points P(-1/3;0), Q(1;0) and R(0;1). Detremine the values of m, n and k.


On the answer sheet they provide more than one approach, one is this:



\(f'(x)=(x+{1 \over3})(x-1)\)


I do understand the +1/3 and the -1, what I do not understand is how is \(mx^2+nx+k\) replaced by \((x+{1 \over3})(x-1)\)


This may sound like a real daft question, but I honestly don't get this..How would I know in a test this is the first step to take?


Please if someone would kindly just put my finite brain to rest...All help is appreciated, thank you all

 Apr 6, 2023

Hi Juriemagic  :)


y is the graph of the first derivate, not of the original function so you can ignore all the derivative notation


Graph  \(y=f'(x)=mx^2+nx+k\)   goes through the points P(-1/3;0), Q(1;0) and R(0;1). Detremine the values of m, n and k.



Graph  \(y=mx^2+nx+k\)   goes through the points P(-1/3;0), Q(1;0) and R(0;1). Detremine the values of m, n and k.


There are at most 2 roots becasue the highest power of x is 2

P and Q are the roots because the y values are 0   so you have 


\(y=t(x+\frac{1}{3})(x-1)\)      where t is some constant.


Now you use the 3rd point  R(0,1) to find  t


\(1=t(0+\frac{1}{3})(0-1)\\ 1=t(\frac{1}{3})(-1)\\ 1=t*-\frac{1}{3}\\ t=-3 \)


So the function is     \(y=-3(x+\frac{1}{3})(x-1)\)


Now expand that our to find  m, n  and k


At the end substitute P, Q and R in to check it is correct.   If not go find your, or mine, error.  

 Apr 6, 2023

Hello Melody!!,


Thank you for the explanation...yes, I am able to take everything further, it was just that step that threw me off. The answer I was looking for really, you gave to me, and it was that the y is the graph of the first derivative, which means it really is Y=mx^2+mx+k.


Okay, I fully have it now. So the formula we used was the y=m(x-x1)(x-x2)...Thank you..stay blessed!!smiley

juriemagic  Apr 7, 2023

Hi Melody,


just a question please?


How would the sum be done If it was \(F'(x)=mx^2+nx+k\) ?

juriemagic  Apr 7, 2023

Hi Juriemagic.


You do not have enought points to get an exact answer for your senario.

(-1/3;0), Q(1;0) and R(0;1)

f(x) is a cubic so we need another point, preferably another root


\(f'(x)=mx^2+nx+k\\ integrate\\ f(x)=\frac{m}{3}x^3+\frac{n}{2}x^2+kx+t\\ y=\frac{m}{3}x^3+\frac{n}{2}x^2+kx+t\\ \)

Sub in the 3 points and you can get all the eunknown constants in terms of just one on them.





\(y=g(x+\frac{1}{3})(x-1)(x-a)\\ sub \;in\; (0,1) \\ 1=g(+\frac{1}{3})(-1)(-a)\\ 1=g(\frac{a}{3})\\ g=\frac{3}{a}\\ so\\ y=\frac{3}{a}(x+\frac{1}{3})(x-1)(x-a)\\ \)


Note :  I have not checked this for careless error. 

 Apr 8, 2023

Hi Melody,


nope that's great thanks...so you would then have to integrate...yes, I agree about the additional info one would need to do this integration. Thanks Melody, your time is always appreciated!..stay blessed..

juriemagic  Apr 9, 2023

Always a pleasure  :)

Melody  Apr 9, 2023

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