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A parabola opening up or down has vertex (0,−5)(0,-5) and passes through (−8,11)(-8,11). Write its equation in vertex form.\newlineSimplify any fractions.

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Q. A parabola opening up or down has vertex (0,−5)(0,-5) and passes through (−8,11)(-8,11). 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,−5)(0, -5)?\newlineSubstitute 00 for hh and −5-5 for kk in the vertex form.\newliney=a(x−0)2+(−5)y = a(x - 0)^2 + (-5)\newliney=ax2−5y = ax^2 - 5
  3. Find 'a' Value: Use the point (−8,11)(-8, 11) to find the value of 'a'.\newlineReplace the variables with (−8,11)(-8, 11) in the equation.\newlineSubstitute −8-8 for xx and 1111 for yy.\newline11=a(−8)2−511 = a(-8)^2 - 5\newline11=64a−511 = 64a - 5
  4. Solve for 'a': Solve for 'a'.\newlineAdd 55 to both sides of the equation.\newline11+5=64a11 + 5 = 64a\newline16=64a16 = 64a\newlineDivide both sides by 6464 to solve for 'a'.\newline1664=a\frac{16}{64} = a\newline14=a\frac{1}{4} = a
  5. Final Equation: Write the equation of the parabola with the value of 'a'.\newlineSubstitute 14\frac{1}{4} for aa in the equation y=ax2−5y = ax^2 - 5.\newliney=(14)x2−5y = \left(\frac{1}{4}\right)x^2 - 5\newlineThe vertex form of the parabola is y=(14)x2−5y = \left(\frac{1}{4}\right)x^2 - 5.

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