The function provided is:

$f(x) = -2x^2 + 20x - 49$

Since the coefficient of $x^2$ (which is $-2$) is negative, the parabola opens downward. Therefore, this function has a maximum value rather than a minimum.

Steps to find the maximum value:

1. The vertex of a quadratic function $f(x) = ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$.

2. Here, $a = -2$ and $b = 20$.

   $x = -\frac{20}{2 \times -2} = \frac{20}{4} = 5$

3. Substitute $x = 5$ back into the function to find the maximum value:

   $f(5) = -2(5)^2 + 20 \cdot 5 - 49$ $f(5) = -2 \cdot 25 + 100 - 49$ $f(5) = -50 + 100 - 49 = 1$

Answers:

• Does the function have a minimum or maximum value? Maximum
• What is the function's maximum value? 1
• Where does the maximum value occur? $x = 5$

Would you like further details on this process, or do you have any questions?

Related Questions:

1. How would the graph change if the coefficient of $x^2$ was positive?
2. How can we identify if a quadratic function has a minimum or maximum without calculating the vertex?
3. What is the importance of the vertex in the context of a quadratic function?
4. How can we rewrite the function in vertex form?
5. What would be the effect on the maximum value if the constant term changed?

Tip:

When dealing with quadratic functions, remember that a negative $a$-value (the coefficient of $x^2$) means the parabola opens downward, leading to a maximum value, while a positive $a$-value indicates a minimum.