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A parabola opening up or down has vertex (0,3)(0,3) and passes through (−8,7)(-8,7). Write its equation in vertex form.\newlineSimplify any fractions.\newline______

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Q. A parabola opening up or down has vertex (0,3)(0,3) and passes through (−8,7)(-8,7). Write its equation in vertex form.\newlineSimplify any fractions.\newline______
  1. Vertex Form of Parabola: 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. Equation with Vertex: What is the equation of a parabola with a vertex at (0,3)(0, 3)?\newlineSubstitute 00 for hh and 33 for kk in the vertex form.\newliney=a(x−0)2+3y = a(x - 0)^2 + 3\newliney=ax2+3y = ax^2 + 3
  3. Find 'a' Using Point: Use the point (−8,7)(-8, 7) to find the value of 'a'.\newlineReplace the variables with (−8,7)(-8, 7) in the equation.\newlineSubstitute −8-8 for xx and 77 for yy.\newline7=a(−8)2+37 = a(-8)^2 + 3\newline7=64a+37 = 64a + 3
  4. Solve for 'a': Solve for 'a'.\newline7=64a+37 = 64a + 3\newlineSubtract 33 from both sides.\newline4=64a4 = 64a\newlineDivide both sides by 6464.\newline464=a\frac{4}{64} = a\newlineSimplify the fraction.\newline116=a\frac{1}{16} = a
  5. Write Equation with 'a': Write the equation of the parabola with the value of 'a'.\newlineSubstitute 116\frac{1}{16} for aa in the equation y=ax2+3y = ax^2 + 3.\newliney=(116)x2+3y = \left(\frac{1}{16}\right)x^2 + 3\newlineVertex form of the parabola: y=(116)x2+3y = \left(\frac{1}{16}\right)x^2 + 3

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