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A parabola opening up or down has vertex (0,−3)(0,-3) and passes through (−6,0)(-6,0). 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 (−6,0)(-6,0). 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. Vertex at (0,−3)(0, -3): What is the equation of a parabola with a vertex at (0,−3)(0, -3)?\newlineSince the vertex is (0,−3)(0, -3), we substitute h=0h = 0 and k=−3k = -3 into the vertex form equation.\newliney=a(x−0)2−3y = a(x - 0)^2 - 3\newliney=ax2−3y = ax^2 - 3
  3. Value of 'a' Calculation: Determine the value of 'a' using the point (−6,0)(-6, 0).\newlineWe know the parabola passes through the point (−6,0)(-6, 0), so we substitute x=−6x = -6 and y=0y = 0 into the equation to find 'a'.\newline0=a(−6)2−30 = a(-6)^2 - 3\newline0=36a−30 = 36a - 3
  4. Solving for 'a': Solve for 'a'.\newlineAdd 33 to both sides of the equation to isolate the term with 'a'.\newline3=36a3 = 36a\newlineDivide both sides by 3636 to solve for 'a'.\newlinea=336a = \frac{3}{36}\newlinea=112a = \frac{1}{12}
  5. Final Equation in Vertex Form: Write the equation of the parabola in vertex form using the value of 'a'.\newlineSubstitute a=112a = \frac{1}{12} into the equation y=ax2−3y = ax^2 - 3.\newliney=(112)x2−3y = \left(\frac{1}{12}\right)x^2 - 3\newlineThis is the equation of the parabola in vertex form.

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