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A parabola opening up or down has vertex (0,−1)(0,-1) and passes through (8,−5)(8,-5). Write its equation in vertex form.\newlineSimplify any fractions.

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Q. A parabola opening up or down has vertex (0,−1)(0,-1) and passes through (8,−5)(8,-5). 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,−1)(0, -1)?\newlineSubstitute 00 for hh and −1-1 for kk in the vertex form.\newliney=a(x−0)2+(−1)y = a(x - 0)^2 + (-1)\newliney=ax2−1y = ax^2 - 1
  3. Use Point to Find 'a': Use the point (8,−5)(8, -5) to find the value of 'a'.\newlineReplace the variables with (8,−5)(8, -5) in the equation.\newlineSubstitute 88 for xx and −5-5 for yy.\newline−5=a(8)2−1-5 = a(8)^2 - 1\newline−5=64a−1-5 = 64a - 1
  4. Solve for 'a': Solve for aa. Add 11 to both sides of the equation to isolate the term with 'a'. −5+1=64a−1+1-5 + 1 = 64a - 1 + 1 −4=64a-4 = 64a Divide both sides by 6464 to solve for 'a'. −4/64=a-4/64 = a −1/16=a-1/16 = a
  5. Write Equation in Vertex Form: Write the equation of the parabola in vertex form using the value of aa. Substitute −116-\frac{1}{16} for aa in the equation y=ax2−1y = ax^2 - 1. y=(−116)x2−1y = \left(-\frac{1}{16}\right)x^2 - 1 The vertex form of the parabola is y=−116x2−1y = -\frac{1}{16}x^2 - 1.

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