Quadratic Function Maximum and Range
The given function is \( y = -x^2 + 4x + 8 \).
This is a quadratic equation in standard form \( y = ax^2 + bx + c \) with a = -1, b = 4, and c = 8.
Since the coefficient 'a' is negative (-1), the parabola opens downwards which means the vertex represents the maximum point on the graph of the function.
The x-coordinate of the vertex can be found by using the formula \( x = -\frac{b}{2a} \).
For the given quadratic function, \( x = -\frac{4}{2 \cdot (-1)} = -\frac{4}{-2} = 2 \).
Now, to find the maximum y-value (which will be the range's upper limit since the parabola opens downwards), we substitute x = 2 back into the original equation.
\( y = -2^2 + 4 \cdot 2 + 8 \)
\( y = -4 + 8 + 8 \)
\( y = 4 + 8 \)
\( y = 12 \)
Therefore, the vertex of the parabola is at (2, 12), and since the parabola opens downwards, the range of y is all real numbers that are less than or equal to 12.
The range is \( y \leq 12 \).
The correct answer is B. \( y \leq 12 \).
