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y=-16x^(2)+180 x+93

y=−16x2+180x+93 y=-16 x^{2}+180 x+93

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Full solution

Q. y=−16x2+180x+93 y=-16 x^{2}+180 x+93
  1. Identify coefficients: Identify the coefficients of the quadratic equation.\newlineGiven y=−16x2+180x+93y = -16x^2 + 180x + 93, the coefficients are a=−16a = -16, b=180b = 180, and c=93c = 93.
  2. Calculate x-coordinate: Calculate the x-coordinate of the vertex using the formula x=−b2ax = -\frac{b}{2a}.\newlinex=−1802×−16=−180−32=5.625x = -\frac{180}{2 \times -16} = -\frac{180}{-32} = 5.625.
  3. Substitute xx into equation: Substitute x=5.625x = 5.625 into the equation to find the y-coordinate of the vertex.\newliney=−16(5.625)2+180(5.625)+93y = -16(5.625)^2 + 180(5.625) + 93\newline=−16(31.640625)+1012.5+93= -16(31.640625) + 1012.5 + 93\newline=−506.25+1012.5+93= -506.25 + 1012.5 + 93\newline=599.25.= 599.25.

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