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A parabola opening up or down has vertex (0,1)(0,-1) and passes through (12,11)(12,11). Write its equation in vertex form.\newlineSimplify any fractions.

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Q. A parabola opening up or down has vertex (0,1)(0,-1) and passes through (12,11)(12,11). Write its equation in vertex form.\newlineSimplify any fractions.
  1. Identify Vertex Values: We have:\newlineVertex: (0,1)(0,-1)\newlineIdentify the values of hh and kk.\newlineVertex is (0,1)(0,-1).\newlineh=0h = 0\newlinek=1k = -1
  2. Select Equation with Substitution: We have:\newliney=a(xh)2+ky = a(x - h)^2 + k\newlineh=0h = 0 and k=1k = -1\newlineSelect the equation after substituting the values of hh and kk.\newlineSubstitute h=0h = 0 and k=1k = -1 in y=a(xh)2+ky = a(x - h)^2 + k.\newliney=a(x0)21y = a(x - 0)^2 - 1\newliney=ax21y = ax^2 - 1
  3. Find Value of a: We have: y=ax21y = ax^2 - 1
    Point: (12,11)(12,11)
    Find the value of a.
    y=ax21y = ax^2 - 1
    11=a(12)2111 = a(12)^2 - 1
    11+1=a×14411 + 1 = a \times 144
    12=a×14412 = a \times 144
    12144=(a×144)144\frac{12}{144} = \frac{(a \times 144)}{144}
    a=112a = \frac{1}{12}
  4. Write Parabola Equation in Vertex Form: We found:\newlinea=112a = \frac{1}{12}\newlineh=0h = 0\newlinek=1k = -1\newlineWrite the equation of a parabola in vertex form.\newliney=a(xh)2+ky = a(x - h)^2 + k\newliney=112(x0)21y = \frac{1}{12}(x - 0)^2 - 1\newlineVertex form: y=112x21y = \frac{1}{12}x^2 - 1

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