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A parabola opening up or down has vertex (0,0)(0,0) and passes through (6,3)(-6,-3). Write its equation in vertex form.\newlineSimplify any fractions.

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Q. A parabola opening up or down has vertex (0,0)(0,0) and passes through (6,3)(-6,-3). Write its equation in vertex form.\newlineSimplify any fractions.
  1. Vertex Form: 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 (0,0)(0,0): 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 (6,3)(-6,-3): y=ax2y = ax^2\newlineReplace the variables with (6,3)(-6, -3) in the equation.\newlineSubstitute 6-6 for xx and 3-3 for yy.\newline3=a(6)2-3 = a(-6)^2\newline3=36a-3 = 36a
  4. Solve for aa: 3=36a-3 = 36a\newlineSolve for aa.\newline336=a-\frac{3}{36} = a\newline112=a-\frac{1}{12} = a
  5. Equation with a=112a = -\frac{1}{12}: y=ax2y = ax^2\newlineWhat is the equation of 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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