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A parabola opening up or down has vertex (0,−3)(0, -3) and passes through (4,−5)(4, -5). Write its equation in vertex form.\newlineSimplify any fractions.\newline______

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Q. A parabola opening up or down has vertex (0,−3)(0, -3) and passes through (4,−5)(4, -5). Write its equation in vertex form.\newlineSimplify any fractions.\newline______
  1. Identify vertex form: Identify the vertex form of a 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. Plug vertex coordinates: Plug the vertex coordinates into the vertex form.\newlineSince the vertex is given as (0,−3)(0, -3), we substitute h=0h = 0 and k=−3k = -3 into the vertex form equation to get y=a(x−0)2−3y = a(x - 0)^2 - 3, which simplifies to y=ax2−3y = ax^2 - 3.
  3. Use point to find 'a': Use the point (4,−5)(4, -5) to find the value of 'a'.\newlineWe know the parabola passes through the point (4,−5)(4, -5), so we substitute x=4x = 4 and y=−5y = -5 into the equation y=ax2−3y = ax^2 - 3 to find 'a'.\newline−5=a(4)2−3-5 = a(4)^2 - 3\newline−5=16a−3-5 = 16a - 3
  4. Solve for 'a': Solve for 'a'.\newlineWe will isolate 'a' by adding 33 to both sides of the equation and then dividing by 1616.\newline−5+3=16a-5 + 3 = 16a\newline−2=16a-2 = 16a\newlinea=−216a = -\frac{2}{16}\newlinea=−18a = -\frac{1}{8}
  5. Write final equation: Write the final equation of the parabola in vertex form.\newlineNow that we have found aa to be −18-\frac{1}{8}, we substitute it back into the equation y=ax2−3y = ax^2 - 3 to get the final equation of the parabola:\newliney=(−18)x2−3y = (-\frac{1}{8})x^2 - 3

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