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)(x−1)
I do understand the +1/3 and the -1, what I do not understand is how is mx2+nx+k replaced by (x+13)(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
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)(x−1) where t is some constant.
Now you use the 3rd point R(0,1) to find t
1=t(0+13)(0−1)1=t(13)(−1)1=t∗−13t=−3
So the function is y=−3(x+13)(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.
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!!
Hi Melody,
just a question please?
How would the sum be done If it was F′(x)=mx2+nx+k ?
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)(x−1)(x−a)subin(0,1)1=g(+13)(−1)(−a)1=g(a3)g=3asoy=3a(x+13)(x−1)(x−a)
Note : I have not checked this for careless error.
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..