The vertex points of a parabola is given by

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

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

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

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

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Solution

The correct option is B $(\frac{- b}{2 a}, \frac{4 a c - b^{2}}{4 a})$
The vertex points of a parabola is given by $(\frac{- b}{2 a}, \frac{4 a c - b^{2}}{4 a})$ , where a,b,c are coefficent of polynomial.

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Similar questions

If b² – 4ac = 0 then The roots of the Quadratic equation ax² + bx + c = 0 are given by :

Q. Which of the given formulae is the correct one to find out the roots of the quadratic equation

a x^{2} + b x + c = 0

The roots of a quadratic equation $a x^{2} + b x + c = 0$ are given by $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ , provided $b^{2} - 4 a c \geq 0$ .

Q. What are the roots of the quadratic equation

a x^{2} + b x + c = 0 ?

Q. Prove that root of equation

a x^{2} + b x + c = 0

\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

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The vertex points of a parabola is given by

The correct option is B (−b2a,4ac−b24a)The vertex points of a parabola is given by (−b2a,4ac−b24a), where a,b,c are coefficent of polynomial.

The correct option is B $(\frac{- b}{2 a}, \frac{4 a c - b^{2}}{4 a})$
The vertex points of a parabola is given by $(\frac{- b}{2 a}, \frac{4 a c - b^{2}}{4 a})$ , where a,b,c are coefficent of polynomial.