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A parabola opening up or down has vertex (0,−5)(0,-5) and passes through (−4,−3)(-4,-3). 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,−3)(-4,-3). 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(x−h)2+ky = a(x - h)^2 + k, where (h,k)(h, k) is the vertex of the parabola.
  2. Equation at Vertex: What is the equation of a parabola with a vertex at (0,−5)(0, -5)?\newlineSubstitute 00 for hh and −5-5 for kk in the vertex form.\newliney=a(x−0)2+(−5)y = a(x - 0)^2 + (-5)\newliney=ax2−5y = ax^2 - 5
  3. Use Point to Find aa: Use the point (−4,−3)(-4, -3) to find the value of aa. Replace the variables with (−4,−3)(-4, -3) in the equation. Substitute −4-4 for xx and −3-3 for yy. −3=a(−4)2−5-3 = a(-4)^2 - 5 −3=16a−5-3 = 16a - 5
  4. Solve for a: Solve for a.\newline−3=16a−5-3 = 16a - 5\newlineAdd 55 to both sides of the equation.\newline2=16a2 = 16a\newlineDivide both sides by 1616.\newline216=a\frac{2}{16} = a\newlineSimplify the fraction.\newline18=a\frac{1}{8} = a
  5. Write Equation with aa: Write the equation of the parabola with the value of aa. Substitute 18\frac{1}{8} for aa in the equation y=ax2−5y = ax^2 - 5. y=(18)x2−5y = \left(\frac{1}{8}\right)x^2 - 5 Vertex form of the parabola: y=(18)x2−5y = \left(\frac{1}{8}\right)x^2 - 5

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