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A parabola opening up or down has vertex (0,1)(0,1) and passes through (−12,13)(-12,13). Write its equation in vertex form.\newlineSimplify any fractions.

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Q. A parabola opening up or down has vertex (0,1)(0,1) and passes through (−12,13)(-12,13). Write its equation in vertex form.\newlineSimplify any fractions.
  1. Find 'a' Value: Now we need to find the value of 'a'. We know that the parabola passes through the point (−12,13)(-12,13). We can substitute x=−12x = -12 and y=13y = 13 into the equation to find 'a'.\newlineSubstituting the point into the equation, we get:\newline13=a(−12)2+113 = a(-12)^2 + 1
  2. Solve for 'a': Now we solve for 'a'. First, we simplify the equation: 13=a(144)+113 = a(144) + 1
  3. Subtract to Isolate 'a': Next, we subtract 11 from both sides to isolate the term with 'a':\newline13−1=a(144)13 - 1 = a(144)\newline12=144a12 = 144a
  4. Divide to Solve 'a': Now we divide both sides by 144144 to solve for 'a':\newline12144=a\frac{12}{144} = a\newlinea=112a = \frac{1}{12}
  5. Write Final Equation: Now that we have the value of aa, we can write the final equation of the parabola in vertex form:\newliney=(112)(x−0)2+1y = \left(\frac{1}{12}\right)(x - 0)^2 + 1\newlineThis simplifies to:\newliney=(112)x2+1y = \left(\frac{1}{12}\right)x^2 + 1

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