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A parabola opening up or down has vertex (0,0)(0,0) and passes through (8,4)(-8,4). 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 (8,4)(-8,4). 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 (8,4)(-8,4): y=ax2y = ax^2\newlineReplace the variables with (8,4)(-8, 4) in the equation.\newlineSubstitute 8-8 for xx and 44 for yy.\newline4=a(8)24 = a(-8)^2\newline4=64a4 = 64a
  4. Solve for aa: 4=64a4 = 64a\newlineSolve for aa.\newline464=a\frac{4}{64} = a\newline116=a\frac{1}{16} = a
  5. Equation with a=116a=\frac{1}{16}: y=ax2y = ax^2\newlineWhat is the equation of the parabola if a=116a = \frac{1}{16}?\newlineSubstitute 116\frac{1}{16} for aa.\newliney=(116)x2y = \left(\frac{1}{16}\right)x^2\newlineVertex form of parabola: y=(116)x2y = \left(\frac{1}{16}\right)x^2

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