What is the vertex of $y= -3x^2-x-(x-3)^2$ ?

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Answer 1 · 2017-07-18T21:30:57

The vertex of the equation $-3x^2-x-(x-3)^2$ would be at point
$(5/8, -119/16)$

First expand out the $(x-3)^2$ part of the equation into $-3x^2-x-(x^2-6x+9)$

Then get rid of the parenthesis, $-3x^2-x-x^2+6x-9$ and combine like terms

$=> -4x^2+5x-9$
The equation for finding the domain of the vertex is $-b/(2a)$

Therefore the domain of the vertex is $-(5)/(2*-4)=5/8$

Input the domain into the function to get the range

$=> -4(5/8)^2+5(5/8)-9 = -119/16$

Therefore the vertex of the equation is $(5/8, -119/16)$

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