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A parabola opening up or down has vertex (0,−5)(0,-5) and passes through (4,−1)(4,-1). 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,−1)(4,-1). 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)?\newlineSince the vertex is (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,−1)(4, -1) to find the value of 'a'.\newlineWe know the parabola passes through the point (4,−1)(4, -1), so we can substitute x=4x = 4 and y=−1y = -1 into the equation to solve for 'a'.\newline−1=a(4)2−5-1 = a(4)^2 - 5\newline−1=16a−5-1 = 16a - 5
  4. Solve for 'a': Solve for 'a'.\newlineAdd 55 to both sides of the equation to isolate the term with 'a'.\newline−1+5=16a−5+5-1 + 5 = 16a - 5 + 5\newline4=16a4 = 16a\newlineDivide both sides by 1616 to solve for 'a'.\newline416=a\frac{4}{16} = a\newline14=a\frac{1}{4} = a
  5. Write Equation in Vertex Form: Write the equation of the parabola in vertex form using the value of aa. Now that we have found aa to be 14\frac{1}{4}, we can substitute it back into the equation we derived in Step 22. y=(14)x2−5y = \left(\frac{1}{4}\right)x^2 - 5 This is the equation of the parabola in vertex form.

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