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A parabola opening up or down has vertex (0,0)(0,0) and passes through (8,−16)(8,-16). Write its equation in vertex form.\newlineSimplify any fractions.

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Q. A parabola opening up or down has vertex (0,0)(0,0) and passes through (8,−16)(8,-16). Write its equation in vertex form.\newlineSimplify any fractions.
  1. Vertex Form Explanation: 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. 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 equation.\newliney=a(x−0)2+0y = a(x - 0)^2 + 0\newliney=ax2y = ax^2
  3. Use Point (8,−16)(8, -16): Use the point (8,−16)(8, -16) to find the value of aa. We know the parabola passes through the point (8,−16)(8, -16), so we can substitute x=8x = 8 and y=−16y = -16 into the equation to solve for aa. −16=a(8)2-16 = a(8)^2 −16=64a-16 = 64a
  4. Solve for 'a': Solve for 'a'.\newlineDivide both sides of the equation by 6464 to find the value of 'aa'.\newlinea=−16/64a = -16 / 64\newlinea=−1/4a = -1 / 4
  5. Write Equation in Vertex Form: Write the equation of the parabola in vertex form using the value of 'a'.\newlineSubstitute −14-\frac{1}{4} for aa in the equation y=ax2y = ax^2.\newliney=(−14)x2y = (-\frac{1}{4})x^2\newlineThis is the equation of the parabola in vertex form.

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