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A parabola opening up or down has vertex (0,5)(0,5) and passes through (−12,−13)(-12,-13). Write its equation in vertex form.\newlineSimplify any fractions.

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Q. A parabola opening up or down has vertex (0,5)(0,5) and passes through (−12,−13)(-12,-13). Write its equation in vertex form.\newlineSimplify any fractions.
  1. Vertex Form: What is the vertex form of the parabola?\newlineThe vertex form of a parabola is given by the equation y=a(x−h)2+ky = a(x - h)^2 + k, where (h,k)(h, k) is the vertex of the parabola.
  2. Equation at Vertex: What is the equation of a parabola with a vertex at (0,5)(0, 5)?\newlineSubstitute 00 for hh and 55 for kk in the vertex form.\newliney=a(x−0)2+5y = a(x - 0)^2 + 5\newliney=ax2+5y = ax^2 + 5
  3. Use Point to Find 'a': Use the point (−12,−13)(-12, -13) to find the value of 'a'.\newlineReplace the variables with (−12,−13)(-12, -13) in the equation.\newlineSubstitute −12-12 for xx and −13-13 for yy.\newline−13=a(−12)2+5-13 = a(-12)^2 + 5\newline−13=144a+5-13 = 144a + 5
  4. Solve for 'a': Solve for 'a'.\newline−13=144a+5-13 = 144a + 5\newlineSubtract 55 from both sides.\newline−18=144a-18 = 144a\newlineDivide both sides by 144144.\newlinea=−18/144a = -18 / 144\newlinea=−1/8a = -1 / 8
  5. Write Final Equation: Write the equation of the parabola with the found value of 'a'.\newlineSubstitute −18-\frac{1}{8} for aa in the equation y=ax2+5y = ax^2 + 5.\newliney=(−18)x2+5y = \left(-\frac{1}{8}\right)x^2 + 5\newlineVertex form of the parabola: y=−18x2+5y = -\frac{1}{8}x^2 + 5

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