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A parabola opening up or down has vertex (0,0)(0,0) and passes through (12,12)(12,12). Write its equation in vertex form.\newlineSimplify any fractions.\newline______

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Q. A parabola opening up or down has vertex (0,0)(0,0) and passes through (12,12)(12,12). Write its equation in vertex form.\newlineSimplify any fractions.\newline______
  1. Vertex Form of Parabola: What is the vertex form of the parabola?\newlineVertex form of parabola: y=a(xh)2+ky = a(x - h)^2 + k
  2. Equation at Vertex (00, 00): What is the equation of a parabola with a vertex at (0,0)(0, 0)?\newlineSubstitute 00 for hh and 00 for kk in vertex form.\newliney=a(x0)2+0y = a(x - 0)^2 + 0\newliney=ax2y = ax^2
  3. Substitute (12,12)(12, 12): y=ax2y = ax^2\newlineReplace the variables with (12,12)(12, 12) in the equation.\newlineSubstitute 1212 for xx and 1212 for yy.\newline12=a(12)212 = a(12)^2\newline12=144a12 = 144a
  4. Solve for aa: 12=144a12 = 144a Solve for aa. 12144=a\frac{12}{144} = a 112=a\frac{1}{12} = a
  5. Equation with a=112a = \frac{1}{12}: y=ax2y = ax^2\newlineWhat is the equation of the parabola if a=112a = \frac{1}{12}?\newlineSubstitute 112\frac{1}{12} for aa.\newliney=(112)x2y = \left(\frac{1}{12}\right)x^2\newlineVertex form of parabola: y=(112)x2y = \left(\frac{1}{12}\right)x^2

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