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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)=mx2+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)=mx2+nx+k

f(x)=(x+13)(x1)

 

I do understand the +1/3 and the -1, what I do not understand is how is mx2+nx+k replaced by (x+13)(x1)

 

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+118703 
+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)=mx2+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=mx2+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+13)(x1)      where t is some constant.

 

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

 

1=t(0+13)(01)1=t(13)(1)1=t13t=3

 

So the function is     y=3(x+13)(x1)

 

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)=mx2+nx+k ?

juriemagic  Apr 7, 2023
 #4
avatar+118703 
+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)=mx2+nx+kintegratef(x)=m3x3+n2x2+kx+ty=m3x3+n2x2+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+13)(x1)(xa)subin(0,1)1=g(+13)(1)(a)1=g(a3)g=3asoy=3a(x+13)(x1)(xa)

 

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+118703 
0

Always a pleasure  :)

Melody  Apr 9, 2023

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