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A parabola opening up or down has vertex (0,5)(0,5) and passes through (−10,−20)(-10,-20). 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 (−10,−20)(-10,-20). Write its equation in vertex form.\newlineSimplify any fractions.
  1. Equation Simplification: Now we have the equation y=a(x−0)2+5y = a(x - 0)^2 + 5, which simplifies to y=ax2+5y = ax^2 + 5. We need to find the value of 'aa' using the point (−10,−20)(-10, -20) that lies on the parabola.
  2. Substitute and Solve: Substitute x=−10x = -10 and y=−20y = -20 into the equation to find 'a'.\newline−20=a(−10)2+5-20 = a(-10)^2 + 5\newline−20=100a+5-20 = 100a + 5\newlineNow, we solve for 'a'.
  3. Isolate 'a': Subtract 55 from both sides of the equation to isolate the term with 'a'.\newline−20−5=100a-20 - 5 = 100a\newline−25=100a-25 = 100a\newlineNow, divide both sides by 100100 to solve for 'a'.
  4. Find 'a' Value: −25100=a-\frac{25}{100} = a\newline−14=a-\frac{1}{4} = a\newlineWe have found the value of 'a' to be −14-\frac{1}{4}. Now we can write the equation of the parabola in vertex form.
  5. Write Vertex Form: The equation of the parabola in vertex form is y=a(x−h)2+ky = a(x - h)^2 + k. Substituting the values of 'aa', 'hh', and 'kk', we get:\newliney=−14(x−0)2+5y = -\frac{1}{4}(x - 0)^2 + 5\newlineThis simplifies to y=−14x2+5y = -\frac{1}{4}x^2 + 5.

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