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A parabola opening up or down has vertex (0,0)(0,0) and passes through (12,9)(12,9). 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 (12,9)(12,9). 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 with Vertex 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. Determine 'a' Value: Determine the value of 'a' using the point (12,9)(12, 9) that lies on the parabola.\newlineWe substitute x=12x = 12 and y=9y = 9 into the equation y=ax2y = ax^2 to find 'a'.\newline9=a(12)29 = a(12)^2\newline9=144a9 = 144a
  4. Solve for 'a': Solve for 'a'.\newlineDivide both sides of the equation by 144144 to isolate 'a'.\newlinea=9144a = \frac{9}{144}\newlinea=116a = \frac{1}{16}
  5. Write Parabola Equation: Write the equation of the parabola using the value of aa. Substitute a=116a = \frac{1}{16} into the equation y=ax2y = ax^2. y=(116)x2y = \left(\frac{1}{16}\right)x^2 This is the equation of the parabola in vertex form.

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