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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.\newline______

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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.\newline______
  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 at Origin: What is the equation of a parabola with a vertex at (0,0)(0, 0)?\newlineSince the vertex is at the origin (0,0)(0, 0), we substitute h=0h = 0 and k=0k = 0 into the vertex form.\newliney=a(x−0)2+0y = a(x - 0)^2 + 0\newliney=ax2y = ax^2
  3. Value of 'a' Calculation: Determine the value of 'a' using the point (8,4)(8, 4) that lies on the parabola.\newlineWe substitute x=8x = 8 and y=4y = 4 into the equation y=ax2y = ax^2 to find 'a'.\newline4=a(8)24 = a(8)^2\newline4=64a4 = 64a
  4. Solving for 'a': Solve for 'a'.\newlineDivide both sides of the equation by 6464 to isolate 'a'.\newlinea=464a = \frac{4}{64}\newlinea=116a = \frac{1}{16}
  5. Final Equation in Vertex Form: Write the equation of the parabola in vertex form using the value of 'a'.\newlineSubstitute a=116a = \frac{1}{16} into the equation y=ax2y = ax^2.\newliney=(116)x2y = \left(\frac{1}{16}\right)x^2\newlineThis is the equation of the parabola in vertex form.

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