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A parabola opening up or down has vertex (0,3)(0,3) and passes through (−8,−13)(-8,-13). Write its equation in vertex form.\newlineSimplify any fractions.

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Q. A parabola opening up or down has vertex (0,3)(0,3) and passes through (−8,−13)(-8,-13). 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. 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. Finding 'a' Value: Use the point (−8,−13)(-8, -13) to find the value of 'a'.\newlineReplace the variables with (−8,−13)(-8, -13) in the equation.\newlineSubstitute −8-8 for xx and −13-13 for yy.\newline−13=a(−8)2+3-13 = a(-8)^2 + 3\newline−13=64a+3-13 = 64a + 3
  4. Solving for 'a': Solve for aa.−13=64a+3-13 = 64a + 3 Subtract 33 from both sides.−16=64a-16 = 64a Divide both sides by 6464.a=−1664a = -\frac{16}{64} Simplify the fraction.a=−14a = -\frac{1}{4}
  5. Final Parabola Equation: Write the equation of the parabola with the value of aa.\newlineSubstitute −14-\frac{1}{4} for aa in the equation y=ax2+3y = ax^2 + 3.\newliney=(−14)x2+3y = \left(-\frac{1}{4}\right)x^2 + 3\newlineThe vertex form of the parabola is y=−(14)x2+3y = -\left(\frac{1}{4}\right)x^2 + 3.

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