Question

If $f(x) = x^2 + ax + b$ has minima at $x = 3$ whose value is $5$, then the values of $a$ and $b$ are respectively

• A. $-6, -14$
• B. $-6, 14$
• C. 14, $-6$
• D. 6, 14

Hint:

The vertex of any upward-opening parabola $f(x) = x^2 + ax + b$ is its minimum point, located at $x = -\frac{a}{2}$. Given that the minimum is at $x = 3$, we have $-\frac{a}{2} = 3 \implies a = -6$. This immediately eliminates choices (C) and (D)!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a quadratic function $f(x) = x^2 + ax + b$. We are told that this function reaches a local minimum value of $5$ precisely at the position $x = 3$. We need to compute the constants $a$ and $b$.

Step 2: Key Formula or Approach:
At a local minimum point $x = c$, the first derivative of the function must vanish, meaning $f'(c) = 0$. Additionally, the coordinates of the minimum point must satisfy the original curve function equation, so $f(3) = 5$.

Step 3: Detailed Explanation:
First, let's find the first derivative of the function: $$f'(x) = \frac{d}{dx}(x^2 + ax + b) = 2x + a$$ Since the function has a minimum at $x = 3$, substitute $x = 3$ into $f'(x) = 0$: $$2(3) + a = 0 \implies 6 + a = 0 \implies a = -6$$ Now that we have determined $a = -6$, our function becomes: $$f(x) = x^2 - 6x + b$$ We are given that the minimum value of the function is $5$ at this point, which means $f(3) = 5$. Let's substitute $x = 3$ into the equation: $$5 = (3)^2 - 6(3) + b$$ $$5 = 9 - 18 + b$$ $$5 = -9 + b \implies b = 5 + 9 = 14$$ Thus, the values are $a = -6$ and $b = 14$.

Step 4: Final Answer:
The values of $a$ and $b$ are $-6$ and $14$ respectively, matching option (B).