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

Solution

$$f(x) \geq 0$$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

$$f\left(-2\right)=\ -\frac{3}{2}\left(-4\right)^2=24$$

Comments

2 months 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