Question

Let \(f(x)\) be a quadratic polynomial in x such that \(f(x)≥0\) for all real numbers x.If \(f(2)=0\) and \(f(4)=6\),then \(f(−2)\) is equal to

• A. 12
• B. 36
• C. 24
• D. 6

Answer 1

The Correct Option is C

We are given that the quadratic expression is always greater than or equal to 0 for all real values of \( x \). This implies the parabola opens upwards and its minimum value is zero. 

The expression passes through the points \( (2, 0) \) and \( (4, 6) \). Since the value of the expression at \( x = 2 \) is 0, this is the vertex of the parabola.

The general form of a quadratic expression with vertex at \( (h, k) \) is: \[ y = a(x - h)^2 + k \]

Using vertex \( (2, 0) \), the expression becomes: \[ y = a(x - 2)^2 \]

To find \( a \), substitute point \( (4, 6) \): \[ 6 = a(4 - 2)^2 = a \cdot 4 \Rightarrow a = \frac{6}{4} = \frac{3}{2} \]

So the quadratic expression is: \[ y = \frac{3}{2}(x - 2)^2 \]

Now calculate the value at \( x = -2 \): \[ y = \frac{3}{2}(-2 - 2)^2 = \frac{3}{2} \cdot 16 = 24 \]

Final Answer:

When \( x = -2 \), the value of the expression is \( \boxed{24} \).

Answer 2

We are given that \( f(x) \geq 0 \) for all real numbers \( x \). This implies the discriminant \( D \leq 0 \).

Since \( f(2) = 0 \), this means \( x = 2 \) is a root of the function. 

Therefore, the function must be of the form: \[ f(x) = a(x - 2)^2 \] (a perfect square with its minimum at \( x = 2 \))

Also given is: \[ f(4) = 6 \] Substituting: \[ 6 = a(4 - 2)^2 = a(2)^2 = 4a \Rightarrow a = \frac{6}{4} = \frac{3}{2} \]

So the function becomes: \[ f(x) = \frac{3}{2}(x - 2)^2 \]

Now, evaluate \( f(-2) \): \[ f(-2) = \frac{3}{2}(-2 - 2)^2 = \frac{3}{2} \cdot 16 = 24 \]

Final Answer:

\[ \boxed{f(-2) = 24} \]