For a quadratic function in standard form, $y = a x^{2} + b x + c$ , Find the axis of symmetry.

y = \frac{- b}{2 a}

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x = \frac{- b}{2 a}

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y = \frac{b^{2} - 4 a c}{2 a}

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None of these

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Solution

The correct option is B $x = \frac{- b}{2 a}$
$a x^{2} + b x + c = 0$ .......( $1$ )
now on differention w.r.t. $x$
$\frac{d y}{d x} = 2 a x + b$
Now for maxima, minima and for axis of symmetry, put $\frac{d y}{d x} = 0$
$0 = 2 a x + b$
Hence, $x = - \frac{b}{2 a}$
Axis of symmetry of parabola is $x = \frac{- b}{2 a}$

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

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the solutions are

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

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For a quadratic function in standard form, y=ax2+bx+c, Find the axis of symmetry.

The correct option is B x=−b2aax2+bx+c=0 .......(1)now on differention w.r.t. xdydx=2ax+bNow for maxima, minima and for axis of symmetry, put dydx=0 0=2ax+bHence, x=−b2aAxis of symmetry of parabola is x=−b2a

For a quadratic function in standard form, $y = a x^{2} + b x + c$ , Find the axis of symmetry.