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A parabola opening up or down has vertex (0,4)(0,-4) and passes through (6,1)(6,-1). Write its equation in vertex form.\newlineSimplify any fractions.

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Q. A parabola opening up or down has vertex (0,4)(0,-4) and passes through (6,1)(6,-1). Write its equation in vertex form.\newlineSimplify any fractions.
  1. Vertex Form: 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 at Vertex: What is the equation of a parabola with a vertex at (0,4)(0, -4)?\newlineSubstitute 00 for hh and 4-4 for kk in vertex form.\newliney=a(x0)2+(4)y = a(x - 0)^2 + (-4)\newliney=ax24y = ax^2 - 4
  3. Substitute Values: y=ax24y = ax^2 - 4\newlineReplace the variables with (6,1)(6, -1) in the equation.\newlineSubstitute 66 for xx and 1-1 for yy.\newline1=a(6)24-1 = a(6)^2 - 4\newline1=36a4-1 = 36a - 4
  4. Solve for aa: 1=36a4-1 = 36a - 4\newlineSolve for aa.\newline1+4=36a-1 + 4 = 36a\newline3=36a3 = 36a\newline336=a\frac{3}{36} = a\newline112=a\frac{1}{12} = a
  5. Equation with a=112a=\frac{1}{12}: y=ax24y = ax^2 - 4\newlineWhat is the equation of the parabola if a=112a = \frac{1}{12}?\newlineSubstitute 112\frac{1}{12} for aa.\newliney=(112)x24y = \left(\frac{1}{12}\right)x^2 - 4\newlineVertex form of the parabola: y=(112)x24y = \left(\frac{1}{12}\right)x^2 - 4

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