What is the vertex form of #y=-x^2+5x-5#?

1 Answer
Jim G.
May 10, 2018

#y=-(x-5/2)^2+5/4#

Explanation:

#"the equation of a parabola in "color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#

#"where "(h,k)" are the coordinates of the vertex and a"#
#"is a multiplier"#

#"to obtain this form use "color(blue)"completing the square"#

#• " the coefficient of the "x^2" term must be 1"#

#"factor out "-1#

#y=-(x^2-5x+5)#

#• " add/subtract "(1/2"coefficient of the x-term")^2" to"#
#x^2-5x#

#y=-(x^2+2(-5/2)x color(red)(+25/4)color(red)(-25/4)+5)#

#color(white)(y)=-(x-5/2)^2-(-25/4+5)#

#rArry=-(x-5/2)^2+5/4larrcolor(red)"in vertex form"#

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