Let $$f(x)$$ be a quadratic polynomial in $$x$$ such that $$f(x) \geq 0$$ for all real numbers $$x$$. If f(2) = 0 and f( 4) = 6, then f(-2) is equal to
$$f(x) \geq 0$$for all real numbers $$x$$, so D<=0
Since f(2)=0 therefore x=2 is a root of f(x)
Since the discriminant of f(x) is less than equal to 0 and 2 is a root so we can conclude that D=0
Therefore f(x) = $$a\left(x-2\right)^2$$
f(4)=6
or, 6 = $$a\left(x-2\right)^2$$
a= 3/2
$$f\left(-2\right)=\ -\frac{3}{2}\left(-4\right)^2=24$$
DHEERAJ KUMAR YADAV
3 months ago
mb
rushikesh fugare
8 months, 1 week ago
How did u get that idea to keep f(x) as a(x-2)^2 +b why didn't u take ax^2 +bx+c
Gautamdev Chakravarty
3 months ago
the form is taken so as to ensure it will be always greater than or equal to zero