Question:medium

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

Updated On: Jan 15, 2026
  • 12

  • 36

  • 24

  • 6

Show Solution

The Correct Option is C

Solution and Explanation

Given that the quadratic expression is \( \ge 0 \) for all real \( x \), the parabola opens upwards with a minimum value of zero. The expression passes through \( (2, 0) \) and \( (4, 6) \). Since the expression is 0 at \( x = 2 \), this point is the vertex.

The vertex form of a quadratic is \( y = a(x - h)^2 + k \). With vertex \( (2, 0) \), this simplifies to \( y = a(x - 2)^2 \).

To determine \( a \), we use the point \( (4, 6) \): \( 6 = a(4 - 2)^2 = a \cdot 4 \). Thus, \( a = \frac{6}{4} = \frac{3}{2} \).

The quadratic expression is \( y = \frac{3}{2}(x - 2)^2 \).

Evaluating at \( x = -2 \): \( y = \frac{3}{2}(-2 - 2)^2 = \frac{3}{2} \cdot 16 = 24 \).

Final Answer:

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

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