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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: huss48 on April 07, 2010, 10:04:47 pm

Title: help with this question (polynomial functions)
Post by: huss48 on April 07, 2010, 10:04:47 pm: The Questions is:

The points with coordinates (1,1) and (2,5) lie on a parabola with equation of the form y=ax^2+b.

Find the values of a and b.

Please help and thanks you
Title: Re: help with this question (polynomial functions)
Post by: brightsky on April 07, 2010, 10:07:06 pm: Substitute the values for the coordinates into the equation then solve simultaneously.

$y = ax^2 + b$

The parabola goes passed the points (1,1) and (2,5), so the equation must satisfy it.

$1 = a(1)^2 + b$ ...(1)

$5 = a(2)^2 + b$ ...(2)

From (1):

$1 = a + b \implies a = 1 - b$ ...(3)

Substitute (3) into (2):

$5 = (1 - b)(2)^2 + b \implies 5 = 4(1 - b) + b \implies 5 = 4 - 4b + b \implies 5 = 4 - 3b \implies 3b = -1 \implies \boxed{b = -\frac{1}{3}}$

Substitute $b = -\frac{1}{3}$ into (3)

$a = 1 - \left(-\frac{1}{3}\right)$

$a = 1 + \frac{1}{3}$

$\boxed{a = \frac{4}{3}}$
Title: Re: help with this question (polynomial functions)
Post by: huss48 on April 07, 2010, 10:08:32 pm: yep ok but can someone do the working out so i can check later.

thanks
Title: Re: help with this question (polynomial functions)
Post by: superflya on April 07, 2010, 10:10:59 pm: sub in ur first point u get $1=a+b$

sub in 2nd point $5=4a+b$

solve simultaneously... $a=\frac{4}{3}$ & $b=\frac{-1}{3}$
Title: Re: help with this question (polynomial functions)
Post by: brightsky on April 07, 2010, 10:11:06 pm: Edited. Hope it helps. :) (...and that it's right. :p)