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What is the vertex of # y= -(x+1)^2+2x^2-x #?

Answer 1

Ethan Adkins

#(3/2,-13/4)#

#"expand and simplify right side of the equation "#

#y=-(x^2+2x+1)+2x^2-x#

#color(white)(y)=-x^2-2x-1+2x^2-x#

#color(white)(x)=x^2-3x-1larrcolor(blue)"in standard form"#

#"with "a=1,b=-3" and "c=-1#

#"the x-coordinate of the vertex is "#

#•color(white)(x)x_(color(red)"vertex")=-b/(2a)=-(-3)/3=3/2#

#"substitute this value into equation for y-coordinate"#

#y_(color(red)"vertex")=(3/2)^2-3(3/2)-1=-13/4#

#rArrcolor(magenta)"vertex"=(3/2,-13/4)#

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Answer 2

Jace Anderson

The vertex of the function ( y = -(x + 1)^2 + 2x^2 - x ) is (-1, 0).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.