Question

Let \( f(x) = x^2 + ax + \beta \). If \( f \) has a local minimum at \( (2,6) \), then \( f(0) \) is equal to

• A. \(10 \)
• B. \(-6 \)
• C. \(8 \)
• D. \(-8 \)
• E. \(6 \)

Hint:

Use both derivative condition and point value for extrema problems.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A function has a local minimum at a point if two conditions are met: 1. The first derivative at that point is zero (\(f'(x) = 0\)). 2. The point itself lies on the graph of the function (\(f(x) = y\)). We will use these two conditions to find the unknown constants `a` and \(\beta\).
Step 2: Key Formula or Approach:
1. The local minimum is at the point (2, 6). This means when \(x=2\), \(y=f(2)=6\). This gives us one equation. 2. The location of a local minimum for a differentiable function occurs where the first derivative is zero. So, \(f'(2) = 0\). This gives us a second equation. 3. Solve the system of two equations for `a` and \(\beta\). 4. Find the value of \(f(0)\).
Step 3: Detailed Explanation:
1. Use the derivative condition. The function is \(f(x) = x^2 + ax + \beta\). Its derivative is \(f'(x) = 2x + a\). Since there is a local minimum at \(x=2\), we have \(f'(2) = 0\). \[ 2(2) + a = 0 \] \[ 4 + a = 0 \implies a = -4 \] 2. Use the point condition. The point (2, 6) is on the graph, which means \(f(2) = 6\). \[ f(2) = (2)^2 + a(2) + \beta = 6 \] \[ 4 + 2a + \beta = 6 \] We already found that \(a = -4\). Substitute this value into the equation: \[ 4 + 2(-4) + \beta = 6 \] \[ 4 - 8 + \beta = 6 \] \[ -4 + \beta = 6 \implies \beta = 10 \] 3. Find f(0). Now we have the complete function: \(f(x) = x^2 - 4x + 10\). The question asks for \(f(0)\). \[ f(0) = (0)^2 - 4(0) + 10 = 10 \] Note that \(f(0)\) is simply the value of the constant term \(\beta\).
Step 4: Final Answer:
The value of \(f(0)\) is 10.