Question

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

Answer 1

`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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