I Need Help With This Maths Question Please - Quadratic Equations (1 Viewer)

[candytrip]

<img src="http://i37.tinypic.com/2e1f0av.jpg">

Omfg I have no idea how to work this out, all I know is that it's a parabola, please show me how to do it, thanks!

 

[Just.Snaz]

umm.. i don't think it's a parabola..

EDIT: k maybe it is.. i dunno, i don't do general lol

 

[Affinity]

depends on what you want.. if you want a polynomial, it's the lagrange polynomial (check wiki)

t = 11[(x-1)(x-2)(x-3)]/6 - 6[(x-1)(x-2)(x-4)]/2 + 3[(x-1)(x-3)(x-4)]/2 - 2[(x-2)(x-3)(x-4)]/6
works out to be t = x^2 - 2x + 3


if you know it's a quadratic, you have
t = ax^2 +bx + c

in particular:
a + b + c = 2
4a + 2b + c = 3
9a + 3b + c = 6

so
3a + b = 1
5a + b = 3

2a =2
a = 1
b = -2
c = 3

and check that the equation works for x=4, t = 11

 

[PC]

In any case, it's well outside the General Maths course!

Still, if you use a graphic calculator, go into the STAT mode, enter the points and plot them on a scatterplot. Then do a curve fit and choose quadratic. You'll get a = 1, b = -2 and c = 3.

So equation is t = x² – 2x + 3.

 

[candytrip]

In any case, it's well outside the General Maths course!

Still, if you use a graphic calculator, go into the STAT mode, enter the points and plot them on a scatterplot. Then do a curve fit and choose quadratic. You'll get a = 1, b = -2 and c = 3.

So equation is t = x² – 2x + 3.

Lol it's actually a question from my school's last year's Year 10 yearlies,