The coordinates of the vertex of a parabola represented by $y = a x^{2} + b x + c$ is, . Take D as discriminant;

(\frac{b}{2 a}, \frac{- D}{4 a})

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(\frac{b}{2 a}, \frac{D}{4 a})

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(\frac{- b}{2 a}, \frac{D}{4 a})

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(\frac{- b}{2 a}, \frac{- D}{4 a})

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Solution

The correct option is D $(\frac{- b}{2 a}, \frac{- D}{4 a})$
We know that, the coordinates of the vertex of parabola of the form ${(x - h)}^{2} = 4 a (y - k) are (h, k) So, y = {ax}^{2} + bx + c can be written as : \frac{y}{a} = x^{2} + \frac{b}{a} \cdot x + \frac{c}{a} \Rightarrow \frac{y}{a} = x^{2} + 2 \cdot \frac{b}{2 a} \cdot x + {(\frac{b}{2 a})}^{2} - {(\frac{b}{2 a})}^{2} + \frac{c}{a} \Rightarrow \frac{y}{a} = {(x + \frac{b}{2 a})}^{2} - \frac{b^{2}}{4 a^{2}} + \frac{c}{a} \Rightarrow {(x + \frac{b}{2 a})}^{2} = \frac{y}{a} + \frac{b^{2}}{4 a^{2}} - \frac{c}{a} \Rightarrow {(x + \frac{b}{2 a})}^{2} = \frac{y}{a} + \frac{\sqrt{D}}{4 a^{2}} \Rightarrow {(x + \frac{b}{2 a})}^{2} = \frac{1}{a} (y + \frac{\sqrt{D}}{4 a}) Hence, the coordinates of the vertex of the given parabola is (\frac{- b}{2 a}, \frac{- \sqrt{D}}{4 a})$

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