Cubic equation from 4 coordinates

[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:

[Something Else]

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:

When (x,y)=(0,1), substitute into you8r equation, and you get:
1=d

When (x,y)=(1,3):
3=a+b+c+d

When (x,y)=(-1,-1):
-1=-a+b-c+d

Whem(x,y)=(2,11):
11=8a+4b+2c+d

So now you have 4 linear equations of a,b,c, and d.
Simultaneous Solve for their unknowns.

Feel free to correct any sloppy maths that I may have done, as I just woke up.

:)
-Lou

[freswood]

You're so right! Thankyou so much! Only problem is - I'm stuck again :( I've got:

2 = a+b+c
-2 = -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.

[Something Else]

Hint:
Add your first two equations together to get B.
Substitute B into your 3 equations, then you'll find 2 linearly independent equations of a and c.

:)