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A parabola opening up or down has vertex (0,0)(0,0) and passes through (4,4)(4,4). 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 (4,4)(4,4). Write its equation in vertex form.\newlineSimplify any fractions.\newline______
  1. Vertex Form Explanation: What is the vertex form of a parabola?\newlineThe vertex form of a parabola is given by the equation y=a(xh)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 (0,0)(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(x0)2+0y = a(x - 0)^2 + 0\newliney=ax2y = ax^2
  3. Value of 'a' Calculation: Determine the value of 'a' using the point (4,4)(4, 4).\newlineWe know the parabola passes through the point (4,4)(4, 4), so we substitute x=4x = 4 and y=4y = 4 into the equation y=ax2y = ax^2.\newline4=a(4)24 = a(4)^2\newline4=16a4 = 16a
  4. Solving for 'a': Solve for 'a'.\newlineDivide both sides of the equation by 1616 to solve for 'a'.\newline416=a\frac{4}{16} = a\newline14=a\frac{1}{4} = a
  5. Final 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 = \left(\frac{1}{4}\right)x^2\newlineThis is the equation of the parabola in vertex form.

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