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A parabola opening up or down has vertex (0,2)(0,2) and passes through (−6,−1)(-6,-1). 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,−1)(-6,-1). Write its equation in vertex form.\newlineSimplify any fractions.
  1. Vertex Form of Parabola: 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)?\newlineSubstitute 00 for hh and 22 for kk in the vertex form.\newliney=a(x−0)2+2y = a(x - 0)^2 + 2\newliney=ax2+2y = ax^2 + 2
  3. Use Point to Find aa: Use the point (−6,−1)(-6, -1) to find the value of aa. Replace the variables with (−6,−1)(-6, -1) in the equation. Substitute −6-6 for xx and −1-1 for yy. −1=a(−6)2+2-1 = a(-6)^2 + 2 −1=36a+2-1 = 36a + 2
  4. Solve for a: Solve for a.\newline−1=36a+2-1 = 36a + 2\newlineSubtract 22 from both sides.\newline−3=36a-3 = 36a\newlineDivide both sides by 3636.\newlinea=−336a = -\frac{3}{36}\newlineSimplify the fraction.\newlinea=−112a = -\frac{1}{12}
  5. Write Equation in Vertex Form: Write the equation of the parabola in vertex form using the value of aa.\newlineSubstitute −112-\frac{1}{12} for aa in the equation y=ax2+2y = ax^2 + 2.\newliney=(−112)x2+2y = \left(-\frac{1}{12}\right)x^2 + 2\newlineThe vertex form of the parabola is y=−112x2+2y = -\frac{1}{12}x^2 + 2.

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