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

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Q. A parabola opening up or down has vertex (0,−5)(0,-5) and passes through (4,−4)(4,-4). Write its equation in vertex form.\newlineSimplify any fractions.
  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,−5)(0, -5), we substitute h=0h = 0 and k=−5k = -5 into the vertex form equation.\newliney=a(x−0)2−5y = a(x - 0)^2 - 5\newliney=ax2−5y = ax^2 - 5
  3. Use point to find 'a': Use the point (4,−4)(4, -4) to find the value of 'a'.\newlineWe know that the parabola passes through the point (4,−4)(4, -4), so we substitute x=4x = 4 and y=−4y = -4 into the equation to solve for 'a'.\newline−4=a(4)2−5-4 = a(4)^2 - 5\newline−4=16a−5-4 = 16a - 5
  4. Solve for 'a': Solve for 'a'.\newlineAdd 55 to both sides of the equation to isolate the term with 'a'.\newline−4+5=16a−5+5-4 + 5 = 16a - 5 + 5\newline1=16a1 = 16a\newlineDivide both sides by 1616 to solve for 'a'.\newline116=a\frac{1}{16} = a
  5. Write final equation: Write the final equation of the parabola in vertex form.\newlineNow that we have the value of ' extit{a}', we can write the equation of the parabola.\newliney=116x2−5y = \frac{1}{16}x^2 - 5

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