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A parabola opening up or down has vertex (0,−5)(0,-5) and passes through (8,11)(8,11). 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 (8,11)(8,11). Write its equation in vertex form.\newlineSimplify any fractions.
  1. Identify Vertex: Identify the vertex of the parabola.\newlineThe given vertex is (0,−5)(0,-5), which means h=0h = 0 and k=−5k = -5.\newlineThe vertex form of a parabola is y=a(x−h)2+ky = a(x - h)^2 + k.\newlineSubstitute 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
  2. Substitute Values: Use the point (8,11)(8,11) to find the value of ′a′'a'. Substitute x=8x = 8 and y=11y = 11 into the equation y=ax2−5y = ax^2 - 5. 11=a(8)2−511 = a(8)^2 - 5 11=64a−511 = 64a - 5 Add 55 to both sides to isolate the term with ′a′'a'. 11+5=64a11 + 5 = 64a ′a′'a'00 Divide both sides by ′a′'a'11 to solve for ′a′'a'. ′a′'a'33 Simplify the fraction. ′a′'a'44
  3. Find Value of 'a': Write the final equation of the parabola in vertex form.\newlineNow that we have a=14a = \frac{1}{4}, h=0h = 0, and k=−5k = -5, substitute these values into the vertex form equation.\newliney=a(x−h)2+ky = a(x - h)^2 + k\newliney=(14)(x−0)2−5y = \left(\frac{1}{4}\right)(x - 0)^2 - 5\newlineSimplify the equation.\newliney=(14)x2−5y = \left(\frac{1}{4}\right)x^2 - 5

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