# What is the vertex of #y=-(x + 2)^2 - 3x+9#?

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To find the vertex of the quadratic function ( y = -(x + 2)^2 - 3x + 9 ), first, rewrite the function in vertex form ( y = a(x - h)^2 + k ). Then, the vertex is at the point ((h, k)). So, ( h ) is the x-coordinate of the vertex, and ( k ) is the y-coordinate of the vertex.

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To find the vertex of the quadratic function ( y = -(x + 2)^2 - 3x + 9 ), we need to rewrite the function in vertex form ( y = a(x - h)^2 + k ), where ( (h, k) ) represents the coordinates of the vertex.

By completing the square, we can rewrite the given function in vertex form:

[ y = -(x + 2)^2 - 3x + 9 ] [ = -1(x^2 + 4x + 4) - 3x + 9 ] [ = -x^2 - 4x - 4 - 3x + 9 ] [ = -x^2 - 7x + 5 ]

Comparing this with the vertex form ( y = a(x - h)^2 + k ), we see that ( a = -1 ) and ( k = 5 ).

To find ( h ), we use the formula ( h = -\frac{b}{2a} ), where ( b = -7 ):

[ h = -\frac{-7}{2(-1)} ] [ h = \frac{7}{2} ]

Therefore, the vertex of the function is ( \left(\frac{7}{2}, 5\right) ).

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