Let $f(x)$ be 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

Since $f(x) \geq 0$ for all real numbers $x$, the quadratic polynomial never goes below the x-axis. It either stays above the x-axis or just touches it.

Since $f(2) = 0$, the parabola touches the x-axis at exactly one point: $x = 2$.

A quadratic that's always non-negative and touches the x-axis at only one point must have that point as a repeated root (also called a double root).

Therefore, both roots of $f(x)$ are equal to 2.

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Let $f(x) = ax^2 + bx + c$ where $a > 0$ (since the parabola opens upward to stay non-negative).

Since both roots are 2, we can write:

$f(x) = a(x - 2)^2$

Expanding this form:

$f(x) = a(x^2 - 4x + 4) = ax^2 - 4ax + 4a$

Comparing with $f(x) = ax^2 + bx + c$:

Coefficient of $x^2$: $a = a$

Coefficient of $x$: $b = -4a$

Constant term: $c = 4a$

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We know $f(4) = 6$. Substituting:

$f(4) = a(4)^2 + b(4) + c = 6$

Using our relationships $b = -4a$ and $c = 4a$:

$16a + 4(-4a) + 4a = 6$

$16a - 16a + 4a = 6$

$4a = 6$

$a = 1.5$

Therefore:

$a = 1.5$

$b = -4a = -4(1.5) = -6$

$c = 4a = 4(1.5) = 6$

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Now we can calculate $f(-2)$:

$f(-2) = a(-2)^2 + b(-2) + c$

$f(-2) = 1.5(4) + (-6)(-2) + 6$

$f(-2) = 6 + 12 + 6$

$f(-2) = 24$

Therefore, $f(-2) = 24$