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A parabola opening up or down has vertex (0,−5)(0,-5) and passes through (−12,4)(-12,4). 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 (−12,4)(-12,4). Write its equation in vertex form.\newlineSimplify any fractions.
  1. Vertex Form Explanation: 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,−5)(0, -5)?\newlineSubstitute 00 for hh and −5-5 for kk in the vertex form.\newliney=a(x−0)2+(−5)y = a(x - 0)^2 + (-5)\newliney=ax2−5y = ax^2 - 5
  3. Find 'a' Value: Use the point (−12,4)(-12, 4) to find the value of 'a'.\newlineReplace the variables with (−12,4)(-12, 4) in the equation.\newlineSubstitute −12-12 for xx and 44 for yy.\newline4=a(−12)2−54 = a(-12)^2 - 5\newline4=144a−54 = 144a - 5
  4. Solve for 'a': Solve for 'a'.\newline4=144a−54 = 144a - 5\newlineAdd 55 to both sides of the equation.\newline4+5=144a4 + 5 = 144a\newline9=144a9 = 144a\newlineDivide both sides by 144144.\newline9144=a\frac{9}{144} = a\newlineSimplify the fraction.\newline116=a\frac{1}{16} = a
  5. Final Equation: Write the equation of the parabola using the value of aa.\newlineSubstitute 116\frac{1}{16} for aa in the equation y=ax2−5y = ax^2 - 5.\newliney=(116)x2−5y = \left(\frac{1}{16}\right)x^2 - 5\newlineThis is the equation of the parabola in vertex form.

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