To determine the values of $a$ and $b$ so that the function $f(x) = x^2 + ax + b$ has a minimum at $(3,9)$, we need to use the following key facts:

1. The minimum occurs at $x = 3$:

   • The function is a parabola, and the minimum of a parabola occurs at the vertex. For a quadratic function of the form $f(x) = Ax^2 + Bx + C$, the vertex is located at $x = -\frac{B}{2A}$.
   • In our case, $A = 1$ (since the coefficient of $x^2$ is 1) and $B = a$, so the vertex occurs at $x = -\frac{a}{2}$.
   • We are given that the minimum occurs at $x = 3$, so: $-\frac{a}{2} = 3$ Solving for $a$: $a = -6$

2. The minimum occurs at $(3,9)$, which means $f(3) = 9$:

   • We substitute $x = 3$ and $f(3) = 9$ into the function: $f(3) = 3^2 + a(3) + b = 9$ Substituting $a = -6$ into the equation: $9 + (-6)(3) + b = 9$ Simplifying: $9 - 18 + b = 9$ $-9 + b = 9$ Solving for $b$: $b = 18$

Conclusion:

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

• $a = -6$
• $b = 18$

Thus, the function $f(x) = x^2 - 6x + 18$ has a minimum at $(3,9)$.

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Would you like further clarification on any step, or do you have any additional questions?

Here are five related questions that might be helpful:

1. How do you find the vertex of a quadratic function in general?
2. What is the significance of the derivative in determining the minimum of a function?
3. How does the rate of change relate to the slope of the tangent line at the vertex?
4. How can we use the second derivative test to confirm that the point is a minimum?
5. What are the key features of a quadratic function in terms of its graph?

Tip: When determining the vertex of a parabola, remember that the formula $x = -\frac{b}{2a}$ is always valid for any quadratic function $f(x) = ax^2 + bx + c$.