Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an left bracket, x, comma, y, right bracket(x,y) point.y, equals, minus, 2, x, squared, minus, 32, x, minus, 144y=−2x 2 −32x−144
Question
Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an left bracket, x, comma, y, right bracket(x,y) point.y, equals, minus, 2, x, squared, minus, 32, x, minus, 144y=−2x 2 −32x−144
Solution
The equation of the parabola you provided is y = -2x^2 - 32x - 144.
The vertex form of a parabola is y = a(x-h)^2 + k, where (h, k) is the vertex of the parabola.
To find the vertex, we first need to complete the square.
The equation y = -2x^2 - 32x - 144 can be rewritten as y = -2(x^2 + 16x) - 144.
To complete the square, we take half of the coefficient of x, square it and add it inside the bracket. Half of 16 is 8, and 8 squared is 64.
So, y = -2(x^2 + 16x + 64 - 64) - 144 simplifies to y = -2[(x + 8)^2 - 64] - 144.
This simplifies further to y = -2(x + 8)^2 + 128 - 144, which simplifies to y = -2(x + 8)^2 - 16.
So, the vertex of the parabola is (-8, -16).
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