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A parabola opening up or down has vertex (0,3)(0,3) and passes through (8,19)(-8,19). Write its equation in vertex form.\newlineSimplify any fractions.

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Q. A parabola opening up or down has vertex (0,3)(0,3) and passes through (8,19)(-8,19). Write its equation in vertex form.\newlineSimplify any fractions.
  1. Vertex Form: What is the vertex form of the parabola?\newlineThe vertex form of a parabola is given by the equation y=a(xh)2+ky = a(x - h)^2 + k, where (h,k)(h, k) is the vertex of the parabola.
  2. Equation at (0,3)(0, 3): What is the equation of a parabola with a vertex at (0,3)(0, 3)?\newlineSubstitute 00 for hh and 33 for kk in the vertex form.\newliney=a(x0)2+3y = a(x - 0)^2 + 3\newliney=ax2+3y = ax^2 + 3
  3. Use Point (8,19)(-8, 19): Use the point (8,19)(-8, 19) to find the value of aa. Replace the variables with (8,19)(-8, 19) in the equation. Substitute 8-8 for xx and 1919 for yy. 19=a(8)2+319 = a(-8)^2 + 3 19=64a+319 = 64a + 3
  4. Solve for 'a': Solve for 'a'.\newline19=64a+319 = 64a + 3\newlineSubtract 33 from both sides.\newline16=64a16 = 64a\newlineDivide both sides by 6464.\newline1664=a\frac{16}{64} = a\newline14=a\frac{1}{4} = a
  5. Write Equation: Write the equation of the parabola with the value of 'a' found.\newlineSubstitute 14\frac{1}{4} for aa in the equation y=ax2+3y = ax^2 + 3.\newliney=(14)x2+3y = \left(\frac{1}{4}\right)x^2 + 3\newlineVertex form of the parabola: y=(14)x2+3y = \left(\frac{1}{4}\right)x^2 + 3

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