Question

If f(x) is a second degree polynomial such that \(f(x) \ge 0 \forall x \in \mathbb{R}>\), \(f(-3) = 0\) and \(f(0) = 18\) then \(f(3) = \)

• A. 36
• B. 72
• C. 144
• D. 288

Hint:

A quadratic \(f(x)\) that is always non-negative (\(f(x) \ge 0\)) and has a real root at \(x=r\) must be of the form \(f(x) = a(x-r)^2\) with \(a>0\). The point \((r,0)\) is the vertex of the parabola.

Answer 1

The Correct Option is B

We are given that \(f(x)\) is a second-degree polynomial, so its graph is a parabola.
The condition \(f(x) \ge 0\) for all real \(x\) means the parabola is always on or above the x-axis.
The condition \(f(-3) = 0\) means the parabola touches the x-axis at \(x=-3\).
For a parabola to be non-negative and touch the x-axis at a single point, that point must be its vertex.
Therefore, the vertex of the parabola is at \((-3, 0)\).
The equation of a parabola with vertex at \((h, k)\) is given by \(f(x) = a(x-h)^2 + k\).
Substituting the vertex coordinates, we get \(f(x) = a(x - (-3))^2 + 0 = a(x+3)^2\).
We are given that \(f(0) = 18\). We can use this to find the value of \(a\).
\(f(0) = a(0+3)^2 = a(9) = 18\).
Solving for \(a\), we find \(a = 18/9 = 2\).
So, the polynomial is \(f(x) = 2(x+3)^2\).
Finally, we need to find the value of \(f(3)\).
\(f(3) = 2(3+3)^2 = 2(6^2) = 2(36) = 72\).