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A parabola opening up or down has vertex (0,−5)(0,-5) and passes through (−10,20)(-10,20). 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 (−10,20)(-10,20). Write its equation in vertex form.\newlineSimplify any fractions.
  1. Vertex Form Explanation: What is the vertex form of the parabola?\newlineVertex 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. Use Point to Find 'a': Use the point (−10,20)(-10, 20) to find the value of 'a'.\newlineReplace the variables with (−10,20)(-10, 20) in the equation.\newlineSubstitute −10-10 for xx and 2020 for yy.\newline20=a(−10)2−520 = a(-10)^2 - 5\newline20=100a−520 = 100a - 5
  4. Solve for 'a': Solve for aa.20=100a−520 = 100a - 5Add 55 to both sides of the equation.25=100a25 = 100aDivide both sides by 100100.25100=a\frac{25}{100} = a14=a\frac{1}{4} = a
  5. Write Equation with 'a': Write the equation of the parabola with the value of aa found.\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\newlineVertex form of the parabola: y=(14)x2−5y = \left(\frac{1}{4}\right)x^2 - 5

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