# Cubic equation from 4 coordinates

• Jan 24th 2006, 01:36 AM
freswood
Cubic equation from 4 coordinates
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:
• Jan 24th 2006, 01:41 AM
CaptainBlack
Quote:

Originally Posted by freswood
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 equations
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
• Jan 24th 2006, 02:02 AM
freswood
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.
• Jan 24th 2006, 02:08 AM
CaptainBlack
Quote:

Originally Posted by freswood
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 be

2=a+b+c
-2=-a+b-c
5=4a+2b+c

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
• Jan 24th 2006, 02:27 AM
freswood
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).
• Jan 24th 2006, 03:21 AM
CaptainBlack
Quote:

Originally Posted by freswood
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:

2 = a+b+c
0 = -a+b-c

2+0=(a+b+c)+(-a+b-c)

or:

2=2b,

so b=1.

RonL
• Jan 24th 2006, 11:44 AM
ThePerfectHacker
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 \$\displaystyle y=x^3+x+1\$
Thus, \$\displaystyle (a,b,c,d)=(1,0,1,1)\$
Here is the graph:
• Jan 24th 2006, 12:36 PM
freswood
Thankyou so much! I really appreciate your help :)