How do you write #y=x^2-8x+20# into vertex form?

2 Answers
Apr 11, 2018

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

Explanation:

#y=[x^2-8x]+20#
#y=[(x-4)^2-16]+20#
#y=(x-4)^2-16+20#
#y=(x-4)^2+4#

Apr 11, 2018

#y=(x-4)^2+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 the method of "color(blue)"completing the square"#

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

#• " add/subtract "(1/2"coefficient of the x-term")^2" to"#
#x^2-8x#

#rArry=x^2+2(-4)xcolor(red)(+16)color(red)(-16)+20#

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