How do you write #y = x^2 + 6x + 10# in vertex form?

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Answer 1 · 2016-08-14T11:45:08

#color(green)(y=(x+3)^2+1)#

#y=x^2+6x+10#

#y=ax^2+bx+c" This is the standard form"#

#"The vertex form is expressed as "#

#y=a(x-h)^2-k#

#F=(h,k)#

#F=(h,k) "represents the focus coordinates"#

#"Solution -1:"#

#y=x^2+6x+9+1#

#color(green)(y=(x+3)^2+1)#

#h=-3#

#k=1#

#"Solution-2:"#

#h=-b/(2a)=-6/(2*1)=-3#

#k=(-3)^2+6*(-3)+10=9-18+10=1#

#a=1#

#"The vertex form:"#

#y=1*(x+3)^2+1#

#color(green)(y=(x+3)^2+1)#