The function given is:

$f(x) = 3(x - 4)^2 - 7$

This is a quadratic function in vertex form: $f(x) = a(x - h)^2 + k$, where $a$, $h$, and $k$ are constants. For this function:

• $a = 3$
• $h = 4$
• $k = -7$

Step 1: Determine if the function has a minimum or maximum

Since $a = 3$ (a positive number), the parabola opens upwards. This means the function has a minimum value.

Step 2: Find the minimum value of the function

For a function in the form $f(x) = a(x - h)^2 + k$, the minimum (or maximum) value occurs at $x = h$. Here, $h = 4$.

Now substitute $x = 4$ into the function to find the minimum value:

$f(4) = 3(4 - 4)^2 - 7 = 3 \cdot 0 - 7 = -7$

Conclusion

The function has a minimum value of $-7$, which occurs at $x = 4$.

Would you like more details or have any questions?

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1. How do you determine whether a quadratic function has a minimum or maximum value?
2. Why does the sign of $a$ in $f(x) = ax^2 + bx + c$ affect the parabola's direction?
3. How do we find the vertex of a quadratic function in standard form?
4. How can we rewrite quadratic functions in vertex form from standard form?
5. What are real-life examples where minimum or maximum values of a quadratic function are useful?

Tip: The vertex form of a quadratic function makes it easy to identify the maximum or minimum value directly.