<p>To locate minimum Q, the coordinates of vertex \( Q \) of the parabola \( f(x) = ax^2 + bx + c \) can be found by using the vertex formula \( x = -\frac{b}{2a} \). The given equation can be rewritten as \( f(x) = -x^2 + 5 \), therefore we have \( a = -1 \) and \( b = 0 \).</p> <p>\[ x_Q = -\frac{b}{2a} = -\frac{0}{2(-1)} = 0 \]</p> <p>To find the y-coordinate of Q, substitute \( x_Q \) into \( f(x) \),</p> <p>\[ y_Q = f(x_Q) = -x_Q^2 + 5 = -(0)^2 + 5 = 5 \]</p> <p>Thus the coordinates of Q are \( (0,5) \).</p> <p>Given that the parabola opens downward (\( a < 0 \)), the vertex is the maximum point P. Therefore, the maximum point P is also at \( (0,5) \).</p>
<p>The image does not provide the necessary information or functions to calculate the specific coordinates or the maximum point as requested in the questions (a) and (b). Therefore, a general solution or approach cannot be provided without additional context or data from the original math problem.</p>
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