The vertex of the parabola $y = a x^{2} + b x + c$ is

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

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

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Solution

The correct option is D $(\frac{- b}{2 a}, \frac{{4 a c - b}^{2}}{4 a})$
Given,

$y = a x^{2} + b x + c$

the above equation can be re-written as

$y - (c - \frac{b^{2}}{4 a}) = a {(x + \frac{b}{2 a})}^{2}$

$\Rightarrow y = a x^{2}$ standard form

Hence the vertex is given by,

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

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

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Similar questions

Q. What are the roots of the quadratic equation

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

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

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

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. 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 of the parabola y=ax2+bx+c is

The correct option is D (−b2a,4ac−b24a)Given,y=ax2+bx+cthe above equation can be re-written asy−(c−b24a)=a(x+b2a)2⇒y=ax2 standard formHence the vertex is given by,(−b2a,c−b24a)∴(−b2a,4ac−b24a)

The vertex of the parabola $y = a x^{2} + b x + c$ is