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There's this question in my textbook that I can't seem to answer:
A cubic function of the form y=ax^3+bx^2+cx+d passes through the points (0,1) (1,3) (-1,-1), (2,11). Find the values of a,b,c, and d.
These are seemingly random points. I don't understand how you could find an equation using them. Maybe I'm just being silly :confused:
Plug the values into the equation and you will have four linear equationsQuote:
Originally Posted by freswood
in a, b, c and d. These can then be solved by the usual methods ( lets
say Gaussian elimination for example.)
For instance:
The first point (0,1) corresponds to x=0, y=1, so:
1=d;
the second point (1,3) corresponds to x=1, y=3, so:
3=a+b+c+d;
the third point (-1,-1) corresponds to x=-1, y=-1, so:
-1=-a+b-c+d;
the fourth point (2,11) corresponds to x=2, y=11, so:
11=8a+4b+2c+d.
RonL
You're so right! Thankyou so much! Only problem is - I'm stuck again :( I've got:
2 = a+b+c
0 = -a+b-c
5= 4a+2b+c
It's no good using elimination with the top two then using the bottom, because that just re-introduces another unknown. Arrrr I'm so sorry for being so stupid.
these should beQuote:
Originally Posted by freswood
2=a+b+c
-2=-a+b-c
5=4a+2b+c
Adding the first two gives:
0=2b,
so b=0, then we have:
-2=-a-c
5=4a+c
which looks like it has solution a=1, c=1.
RonL
Thanks so much for your help! Could you just clarify what you meant by "adding the first two" because I can't seem to do it by elimination (ie can I see your working out? :D ) I just made them equal each other (by first making the second one a +2).
If:Quote:
Originally Posted by freswood
2 = a+b+c
0 = -a+b-c
adding these (meaning adding the RHSs and LHSs) gives:
2+0=(a+b+c)+(-a+b-c)
or:
2=2b,
so b=1.
RonL
Just to add, this is a type of problem of drawing a 3rd degreed polynomial of best fist through these points. It happens to be that it is possible to draw a perfect polynomial through these points.
Its equation is
Thus,
Here is the graph:
Thankyou so much! I really appreciate your help :)