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

Question

1280 views around the world

You can reuse this answer
Creative Commons License

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)#