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RolyPoly Group Title

Find vertex and axis of symmetry from the quadratic equation \(y=ax^2+bx+c\) Also, derive the quadratic formula.

  • one year ago
  • one year ago

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  1. RolyPoly Group Title
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    \[y=ax^2 + bx + c \]\[y = a(x^2+\frac{b}{a}x)+c\]\[y=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2-(\frac{b}{2a})^2)+c\]\[y=a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c\]\[y=a(x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a}\] So, vertex is at \(( -\frac{b}{2a} , -\frac{b^2-4ac}{4a})\), axis of symmetry is x=\( -\frac{b}{2a}\)

    • one year ago
  2. RolyPoly Group Title
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    Deriving quadratic formula To find x-intercepts, we put y=0 \[ax^2 + bx + c=0\]\[x^2 + \frac{b}{a}x + \frac{c}{a}=0\]\[x^2 + \frac{b}{a}x + (\frac{b}{2a})^2-(\frac{b}{2a})^2+\frac{c}{a}=0\]\[(x+\frac{b}{2a})^2-(\frac{b^2-4ac}{4a^2})=0\]\[(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\]\[x+\frac{b}{2a}=\pm \sqrt{\frac{b^2-4ac}{4a^2}}\]\[x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}\]\[x=\pm \frac{\sqrt{b^2-4ac}}{2a}-\frac{b}{2a}\]\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

    • one year ago
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