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A parabola opening up or down has vertex (0,4)(0,4) 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 (0,4)(0,4) and passes through (8,−4)(8,-4). Write its equation in vertex form.\newlineSimplify any fractions.
  1. Vertex Form Explanation: What is the vertex form of the parabola?\newlineVertex 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 with Vertex: What is the equation of a parabola with a vertex at (0,4)(0, 4)?\newlineSubstitute 00 for hh and 44 for kk in the vertex form.\newliney=a(x−0)2+4y = a(x - 0)^2 + 4\newliney=ax2+4y = ax^2 + 4
  3. Use Point for 'a': Use the point (8,−4)(8, -4) to find the value of 'a'.\newlineReplace the variables with (8,−4)(8, -4) in the equation.\newlineSubstitute 88 for xx and −4-4 for yy.\newline−4=a(8)2+4-4 = a(8)^2 + 4\newline−4=64a+4-4 = 64a + 4
  4. Solve for 'a': Solve for 'a'.\newline−4=64a+4-4 = 64a + 4\newlineSubtract 44 from both sides to isolate the term with 'a'.\newline−4−4=64a+4−4-4 - 4 = 64a + 4 - 4\newline−8=64a-8 = 64a\newlineDivide both sides by 6464 to solve for 'a'.\newline−864=a-\frac{8}{64} = a\newline−18=a-\frac{1}{8} = a
  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+4y = ax^2 + 4.\newliney=(−18)x2+4y = \left(-\frac{1}{8}\right)x^2 + 4\newlineVertex form of the parabola: y=−(18)x2+4y = -\left(\frac{1}{8}\right)x^2 + 4

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