+0  
 
0
67
6
avatar+1124 

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:

 

\(y=f'(x)=mx^2+nx+k\)

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

 

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
 #1
avatar+118623 
+1

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.

becomes

 

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
 #2
avatar+1124 
+1

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
 #3
avatar+1124 
+1

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
 #4
avatar+118623 
+1

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.

OR

 

alternatively

 

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

 

Note :  I have not checked this for careless error. 

 Apr 8, 2023
 #5
avatar+1124 
+1

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
 #6
avatar+118623 
0

Always a pleasure  :)

Melody  Apr 9, 2023

1 Online Users

avatar