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A parabola opening up or down has vertex (0,−5)(0,-5) and passes through (6,−2)(6,-2). Write its equation in vertex form.\newlineSimplify any fractions.\newline______

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Q. A parabola opening up or down has vertex (0,−5)(0,-5) and passes through (6,−2)(6,-2). Write its equation in vertex form.\newlineSimplify any fractions.\newline______
  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. Use Point to Find 'a': Use the point (6,−2)(6, -2) to find the value of 'a'.\newlineReplace the variables with (6,−2)(6, -2) in the equation.\newlineSubstitute 66 for xx and −2-2 for yy.\newline−2=a(6)2−5-2 = a(6)^2 - 5\newline−2=36a−5-2 = 36a - 5
  4. Solve for 'a': Solve for aa.\newlineAdd 55 to both sides of the equation.\newline−2+5=36a-2 + 5 = 36a\newline3=36a3 = 36a\newlineDivide both sides by 3636 to solve for aa.\newline336=a\frac{3}{36} = a\newline112=a\frac{1}{12} = a
  5. Write Equation with 'a': Write the equation of the parabola with the value of aa found.\newlineSubstitute 112\frac{1}{12} for aa in the equation y=ax2−5y = ax^2 - 5.\newliney=(112)x2−5y = \left(\frac{1}{12}\right)x^2 - 5\newlineVertex form of the parabola: y=(112)x2−5y = \left(\frac{1}{12}\right)x^2 - 5

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