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A parabola opening up or down has vertex (0,6)(0,6) and passes through (12,12)(12,-12). Write its equation in vertex form.\newlineSimplify any fractions.

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Q. A parabola opening up or down has vertex (0,6)(0,6) and passes through (12,12)(12,-12). Write its equation in vertex form.\newlineSimplify any fractions.
  1. Vertex form explanation: What is the vertex form of the parabola?\newlineVertex form of a parabola: y=a(xh)2+ky = a(x - h)^2 + k
  2. Equation with vertex: What is the equation of a parabola with a vertex at (0,6)(0, 6)?\newlineSubstitute 00 for hh and 66 for kk in vertex form.\newliney=a(x0)2+6y = a(x - 0)^2 + 6\newliney=ax2+6y = ax^2 + 6
  3. Substitute values: y=ax2+6y = ax^2 + 6\newlineReplace the variables with (12,12)(12, -12) in the equation.\newlineSubstitute 1212 for xx and 12-12 for yy.\newline12=a(12)2+6-12 = a(12)^2 + 6\newline12=144a+6-12 = 144a + 6
  4. Solve for aa: 12=144a+6-12 = 144a + 6\newlineSolve for aa.\newline126=144a-12 - 6 = 144a\newline18=144a-18 = 144a\newlinea=18144a = \frac{-18}{144}\newlinea=18a = \frac{-1}{8}
  5. Equation with a=18a=-\frac{1}{8}: y=ax2+6y = ax^2 + 6\newlineWhat is the equation of the parabola if a=18a = -\frac{1}{8}?\newlineSubstitute 18-\frac{1}{8} for aa.\newliney=(18)x2+6y = \left(-\frac{1}{8}\right)x^2 + 6\newlineVertex form of the parabola: y=(18)x2+6y = -\left(\frac{1}{8}\right)x^2 + 6

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