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

The correct answer is C:24
Given that the quadratic expression is always greater than or equal to 0 for any real number x,it indicates that its graph forms a U-shaped curve opening upwards. 
Given the points (2,0) and (4, 6) on the curve, it means the lowest point (vertex) of the curve is at (2,0). 
The equation of a quadratic expression can be written as \(y=a(x-h)^2+k\),where (h, k) represents the vertex. 
Using the vertex coordinates (2, 0),the quadratic expression takes the form \(y=a(x - 2)^2\). 
Using the fact that y=6 when x=4, we can calculate the value of a: 
\(6=a(4 - 2)^2\) 
6=4a 
\(a=\frac{6}{4}=\frac{3}{2}\) 
So, the quadratic expression becomes \(y = \frac{3}{2} \times (x - 2)^2\). 
Now, let's find the value of the expression when x = -2: 
\(y = \frac{3}{2} \times (-2 - 2)^2\) 
\(y = \frac{3}{2} \times (-4)^2\) 
\(y = \frac{3}{2} \times 16\) 
y=24 
Therefore, when x=-2, the value of the expression is 24. 
The correct answer is option c. 24 

Answer 2

\(f(x)≥0D≤0\)  for all real numbers x.
We may determine that D=0 since 2 is a root and the discriminant of f(x) is less than or equal to 0.
Given that \(f(2)=0, x=2 \) is a root of \(f(x)\).
Consequently, \(f(x) = a(x-2)^ 2\)
When \(f(4)=6\)
\(⇒ 6 =  a(x−2) ^2\)
\(a=\frac{3}{2}\)
\(⇒ f(-2)=-\frac{3}{2}(-4)^2=24\)