To find the minimum value of the quadratic function $f(x) = 3x^2 + 6x + 7$, we follow these steps:

Step 1: Identify the standard form of a quadratic function

The given function is already in the standard form: $f(x) = ax^2 + bx + c$ where $a = 3$, $b = 6$, and $c = 7$.

Step 2: Find the vertex of the parabola

The minimum (or maximum) value of a quadratic function occurs at the vertex. The x-coordinate of the vertex can be found using the formula: $x = \frac{-b}{2a}$

Substitute $b = 6$ and $a = 3$ into the formula: $x = \frac{-6}{2(3)} = \frac{-6}{6} = -1$

Step 3: Calculate the minimum value

Now that we know $x = -1$, we can substitute this value back into the original function to find the minimum value of $f(x)$: $f(-1) = 3(-1)^2 + 6(-1) + 7$ $f(-1) = 3(1) - 6 + 7$ $f(-1) = 3 - 6 + 7 = 4$

Conclusion:

The minimum value of the function $f(x) = 3x^2 + 6x + 7$ is 4, and it occurs at $x = -1$.

Would you like more details or have any other questions?

Here are 5 questions for further practice:

1. How would you find the maximum value of a quadratic function?
2. Can the quadratic function $f(x) = 3x^2 + 6x + 7$ have a maximum value?
3. What happens to the graph of $f(x)$ when the value of $a$ is negative?
4. How would you solve $f(x) = 0$ for this function?
5. What is the axis of symmetry of $f(x) = 3x^2 + 6x + 7$?

Tip: The sign of the coefficient $a$ determines whether a quadratic function opens upwards (minimum point) or downwards (maximum point).