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A parabola opening up or down has vertex (0,4)(0,4) and passes through (−6,1)(-6,1). Write its equation in vertex form.\newlineSimplify any fractions.

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Q. A parabola opening up or down has vertex (0,4)(0,4) and passes through (−6,1)(-6,1). 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. Substitute vertex form: Substitute the vertex into the vertex form.\newlineSince the vertex is given as (0,4)(0, 4), we substitute h=0h = 0 and k=4k = 4 into the vertex form equation.\newliney=a(x−0)2+4y = a(x - 0)^2 + 4\newliney=ax2+4y = ax^2 + 4
  3. Use point to find 'a': Use the point (−6,1)(-6, 1) to find the value of 'a'.\newlineThe parabola passes through the point (−6,1)(-6, 1), so we substitute x=−6x = -6 and y=1y = 1 into the equation to solve for 'a'.\newline1=a(−6)2+41 = a(-6)^2 + 4\newline1=36a+41 = 36a + 4
  4. Solve for 'a': Solve for 'a'.\newlineSubtract 44 from both sides of the equation to isolate the term with 'a'.\newline1−4=36a1 - 4 = 36a\newline−3=36a-3 = 36a\newlineDivide both sides by 3636 to solve for 'a'.\newlinea=−336a = -\frac{3}{36}\newlinea=−112a = -\frac{1}{12}
  5. Write final equation: Write the final equation of the parabola in vertex form.\newlineNow that we have the value of aa, we can write the equation of the parabola.\newliney=(−112)(x−0)2+4y = \left(-\frac{1}{12}\right)(x - 0)^2 + 4\newliney=(−112)x2+4y = \left(-\frac{1}{12}\right)x^2 + 4

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