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A parabola opening up or down has vertex (0,6)(0,6) and passes through (10,19)(10,-19). 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 (10,19)(10,-19). 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(xh)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 66 for kk in the vertex form.\newliney=a(x0)2+6y = a(x - 0)^2 + 6\newliney=ax2+6y = ax^2 + 6
  3. Find 'a' Value: Use the point (10,19)(10, -19) to find the value of 'a'.\newlineReplace the variables with (10,19)(10, -19) in the equation.\newlineSubstitute 1010 for xx and 19-19 for yy.\newline19=a(10)2+6-19 = a(10)^2 + 6\newline19=100a+6-19 = 100a + 6
  4. Solve for 'a': Solve for 'a'.\newline19=100a+6-19 = 100a + 6\newlineSubtract 66 from both sides.\newline196=100a-19 - 6 = 100a\newline25=100a-25 = 100a\newlineDivide both sides by 100100.\newline25100=a-\frac{25}{100} = a\newline14=a-\frac{1}{4} = a
  5. Write Equation with 'a' Value: Write the equation of the parabola with the value of 'a'.\newlineSubstitute 14-\frac{1}{4} for aa in the equation y=ax2+6y = ax^2 + 6.\newliney=(14)x2+6y = \left(-\frac{1}{4}\right)x^2 + 6\newlineVertex form of the parabola: y=(14)x2+6y = -\left(\frac{1}{4}\right)x^2 + 6

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