What is the vertex form of #y= x^2 +8x +16#?

1 Answer
Tony B
Apr 26, 2016

#color(blue)(y=(x+4)^2)#

Explanation:

Consider the standard for #" "y=ax^2+bx+c#

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#color(blue)("Scenario 1:" -> a=1)" "# (as in your question)

Write as
#y=(x^2+bx)+c#

Take the square outside the bracket.
Add a correction constant k ( or any letter you so chose)

#y=(x+bx)^2+c +k#

Remove the #x# from #b x#

#y=(x + b)^2+c+k#

Halve #b#

#y=(x+b/2)^2+c+k#

Set the value of #k =(-1)xx(b/2)^2#

#y=(x+b/2)^2+c-(b/2)^2#

Substituting the value gives:

#y=(x+8/2)^2+16-16#

#color(blue)(y=(x+4)^2)#
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By changing the brackets content so that it has #b/2# and then squaring #b/2# you introduce a value that was not in the original equation. So you remove this using #k# and thus returning the whole to its original inherent value.
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#color(blue)("Scenario 2:" -> a !=1)#

Write as
#y=a(x^2+b/(2a)x)+c+k#

and you end up with

#y=a(x+b/(2a))^2+c+k#

In this case #k=(-1)xx ((ab)/(2a))^2 = -(b/2)^2#

#y=a(x+b/(2a))^2+c-(b/2)^2#
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