Someone please help with this precalc problem!!!
Favorites|Homepage
Subscriptions | sitemap
HOME > Mathematics > Someone please help with this precalc problem!!!

Someone please help with this precalc problem!!!

[From: ] [author: ] [Date: 11-09-06] [Hit: ]
2) Expanding and simplifying the above,3) So, in general the polynomial,As f(3) = 108,......
Find an expression for a cubic function f if f(3) = 108 and f(-3) = f(0) = f(4) = 0.

-
1) As, f(-3) = f(0) = f(4) = 0, {-3, 0, 4} are the roots of the required cubic polynomial. Hence, the cubic polynomial is given by: (x+3)(x-0)(x-4) = 0

2) Expanding and simplifying the above, x³ - x² - 12x = 0

3) So, in general the polynomial, f(x) = a(x³ - x² - 12x)

As f(3) = 108,

==> 108 = a(27 - 9 - 36); solving a = -6

Thus the required polynomial is f(x) = -6(x³ - x² - 12x)

or in another form the same is: f(x) = 6x(12x + x - x²)
1
keywords: please,problem,this,precalc,help,with,Someone,Someone please help with this precalc problem!!!
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .