Question - Quadratic Function Maximum and Range
Solution:
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. $$ y \leq 12 $$.
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