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A parabola opening up or down has vertex (0,−2)(0,-2) and passes through (−6,7)(-6,7). Write its equation in vertex form.\newlineSimplify any fractions.

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Q. A parabola opening up or down has vertex (0,−2)(0,-2) and passes through (−6,7)(-6,7). 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,−2)(0, -2)?\newlineSince the vertex is (0,−2)(0, -2), we substitute h=0h = 0 and k=−2k = -2 into the vertex form equation.\newliney=a(x−0)2−2y = a(x - 0)^2 - 2\newliney=ax2−2y = ax^2 - 2
  3. Value of 'a' Calculation: Determine the value of 'a' using the point (−6,7)(-6, 7).\newlineWe know the parabola passes through the point (−6,7)(-6, 7), so we substitute x=−6x = -6 and y=7y = 7 into the equation to find 'a'.\newline7=a(−6)2−27 = a(-6)^2 - 2\newline7=36a−27 = 36a - 2
  4. Solving for 'a': Solve for 'a'.\newlineTo find the value of 'a', we solve the equation from the previous step.\newline7=36a−27 = 36a - 2\newline7+2=36a7 + 2 = 36a\newline9=36a9 = 36a\newlinea=936a = \frac{9}{36}\newlinea=14a = \frac{1}{4}
  5. Final Equation in Vertex Form: Write the equation of the parabola in vertex form using the value of aa.\newlineNow that we have found aa to be 14\frac{1}{4}, we substitute it back into the equation y=ax2−2y = ax^2 - 2.\newliney=(14)x2−2y = \left(\frac{1}{4}\right)x^2 - 2\newlineThis is the equation of the parabola in vertex form.

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