If the graph of y = ax^2 + bx + c is as follo | vedclass

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(B) Let the coordinates of $A$ be $(-k, 0)$ and $C$ be $(k, 0)$. Since $AC = 4\sqrt{2}$,we have $2k = 4\sqrt{2}$,so $k = 2\sqrt{2}$.Thus,$A = (-2\sqrt

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$-2\sqrt{2}$

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(B) Let the coordinates of $A$ be $(-k, 0)$ and $C$ be $(k, 0)$. Since $AC = 4\sqrt{2}$,we have $2k = 4\sqrt{2}$,so $k = 2\sqrt{2}$.Thus,$A = (-2\sqrt{2}, 0)$ and $C = (2\sqrt{2}, 0)$.Since $\Delta ABC$ is a right-angled isosceles triangle with the right angle at $B$,$B$ must lie on the $y$-axis at $(0, -h)$ where $h > 0$.The midpoint of $AC$ is the origin $(0, 0)$. In a right-angled isosceles triangle,the median to the hypotenuse is half the length of the hypotenuse.Therefore,the length of the median $OB = \frac{1}{2} AC = \frac{1}{2} (4\sqrt{2}) = 2\sqrt{2}$.Since $B$ is below the $x$-axis,its coordinates are $(0, -2\sqrt{2})$.The minimum value of the quadratic $y = ax^2 + bx + c$ is the $y$-coordinate of its vertex $B$.Thus,the minimum value is $-2\sqrt{2}$.

If the graph of $y = ax^2 + bx + c$ is as follows,where $\Delta ABC$ is a right-angled isosceles triangle with hypotenuse $AC = 4\sqrt{2} \text{ units}$,then the minimum value of $ax^2 + bx + c$ is:

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