The problem asks to determine the values of $a$ and $b$ such that the function $f(x) = x^2 + ax + b$ has a minimum at the point $(3, 9)$. Additionally, you are instructed to use the knowledge of the first derivative and second derivative to achieve the solution.

To solve this, let's break it down:

1. First Condition (minimum at $(3, 9)$):

   • The minimum occurs at $x = 3$, so the derivative of $f(x)$, denoted $f'(x)$, must be equal to zero at $x = 3$.
   • The value of the function at $x = 3$ should be 9, i.e., $f(3) = 9$.

2. Second Condition (derivative analysis):

   • Since the function has a minimum at $x = 3$, the second derivative $f''(x)$ should be positive at that point, confirming it's a minimum.

Step-by-step solution:

1. Start by finding the first derivative of $f(x)$:

$f'(x) = 2x + a$

2. Set $f'(3) = 0$ to satisfy the minimum condition:

$2(3) + a = 0 \implies a = -6$

3. Now, find the value of $b$. Since $f(3) = 9$, substitute $x = 3$ into $f(x)$ and solve for $b$:

$f(3) = 3^2 + (-6)(3) + b = 9$ $9 - 18 + b = 9 \implies b = 18$

Thus, $a = -6$ and $b = 18$.

Final answer:

The values of $a$ and $b$ are:

• $a = -6$
• $b = 18$

Would you like any further details or clarifications on this problem? Here are some related questions you could explore:

1. How would the solution change if the function had a maximum at $(3, 9)$ instead of a minimum?
2. What would the function look like graphically with these values for $a$ and $b$?
3. How can you verify the result using the second derivative test?
4. What happens if the function is cubic instead of quadratic? How does that affect the conditions for a minimum?
5. How do you determine the critical points of a general quadratic function?

Tip: When dealing with quadratic functions, always start by checking the first and second derivatives to determine if the critical points are minima, maxima, or saddle points.