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A parabola opening up or down has vertex (0,−6)(0,-6) and passes through (−8,10)(-8,10). Write its equation in vertex form.\newlineSimplify any fractions.

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Q. A parabola opening up or down has vertex (0,−6)(0,-6) and passes through (−8,10)(-8,10). 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,−6)(0, -6)?\newlineSubstitute 00 for hh and −6-6 for kk in the vertex form.\newliney=a(x−0)2+(−6)y = a(x - 0)^2 + (-6)\newliney=ax2−6y = ax^2 - 6
  3. Use Point to Find 'a': Use the point (−8,10)(-8, 10) to find the value of 'a'.\newlineReplace the variables with (−8,10)(-8, 10) in the equation.\newlineSubstitute −8-8 for xx and 1010 for yy.\newline10=a(−8)2−610 = a(-8)^2 - 6\newline10=64a−610 = 64a - 6
  4. Solve for 'a': Solve for aa.10=64a−610 = 64a - 6Add 66 to both sides of the equation.16=64a16 = 64aDivide both sides by 6464.1664=a\frac{16}{64} = a14=a\frac{1}{4} = a
  5. Write Equation with 'a': Write the equation of the parabola with the value of aa. Substitute 14\frac{1}{4} for aa in the equation y=ax2−6y = ax^2 - 6. y=(14)x2−6y = \left(\frac{1}{4}\right)x^2 - 6 Vertex form of the parabola: y=(14)x2−6y = \left(\frac{1}{4}\right)x^2 - 6

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