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Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an (x,y) point. y=x^(2)-6x+9 Answer:

Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an (x,y) (x, y) point.\newliney=x2−6x+9 y=x^{2}-6 x+9 \newlineAnswer:

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Q. Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an (x,y) (x, y) point.\newliney=x2−6x+9 y=x^{2}-6 x+9 \newlineAnswer:
  1. Calculate x-coordinate: To find the vertex of a parabola given by the equation y=ax2+bx+cy = ax^2 + bx + c, we can use the vertex formula x=−b2ax = -\frac{b}{2a} for the x-coordinate of the vertex.\newlineIn our equation, a=1a = 1 and b=−6b = -6.\newlineLet's calculate the x-coordinate of the vertex.\newlinex=−(−6)/(2⋅1)=62=3x = -(-6)/(2\cdot1) = \frac{6}{2} = 3
  2. Calculate y-coordinate: Now that we have the x-coordinate of the vertex, we can substitute it back into the original equation to find the y-coordinate of the vertex.\newliney=(3)2−6×(3)+9y = (3)^2 - 6\times(3) + 9\newlineLet's calculate the y-coordinate.\newliney=9−18+9=0y = 9 - 18 + 9 = 0
  3. Find vertex coordinates: We have found the x-coordinate (3)(3) and the y-coordinate (0)(0) of the vertex. Therefore, the coordinates of the vertex are (3,0)(3, 0).

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