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A parabola opening up or down has vertex (0,−1)(0,-1) and passes through (6,−4)(6,-4). 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 (6,−4)(6,-4). 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)?\newlineSince the vertex is (0,−1)(0, -1), we substitute h=0h = 0 and k=−1k = -1 into the vertex form equation.\newliney=a(x−0)2−1y = a(x - 0)^2 - 1\newliney=ax2−1y = ax^2 - 1
  3. Use Point to Find 'a': Use the point (6,−4)(6, -4) to find the value of 'a'.\newlineThe parabola passes through the point (6,−4)(6, -4), so we substitute x=6x = 6 and y=−4y = -4 into the equation to solve for 'a'.\newline−4=a(6)2−1-4 = a(6)^2 - 1\newline−4=36a−1-4 = 36a - 1
  4. Solve for 'a': Solve for 'a'.\newlineAdd 11 to both sides of the equation to isolate the term with 'a'.\newline−4+1=36a−1+1-4 + 1 = 36a - 1 + 1\newline−3=36a-3 = 36a\newlineDivide both sides by 3636 to solve for 'a'.\newline−336=a-\frac{3}{36} = a\newlineSimplify the fraction.\newline−112=a-\frac{1}{12} = a
  5. Write Equation in Vertex Form: Write the equation of the parabola in vertex form using the value of aa. Substitute −112-\frac{1}{12} for aa in the equation y=ax2−1y = ax^2 - 1. y=(−112)x2−1y = \left(-\frac{1}{12}\right)x^2 - 1 This is the equation of the parabola in vertex form.

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