Question

How do you write `f(x) = -x^2 + 6x + 8` in vertex form?

Answer 1

`" "y=-(x-3)^2+17`
Or if you prefer: `y=(-x+3)^2+17`

Explanation:

This process introduces an error that has to be compensated for by the inclusion of the correction factor of `k`. This has to be so to maintain the truth of the expression being equal that of y.

In that `f(x)=-x^2+6x+8+k` at this stage `k=0`

Write as:`y=( -1x^2+6x)+8+k`

Factor out the `-1` (Sometimes this value is not 1)

`y=-1(x^2-6x)+8+k`

take the exponent (power of 2) outside the bracket

`y=-1(x-6x)^2+8+k" "larr" now "k" comes into effect."`

Note that each step the value of `k` changes until we finish modifying at which point it becomes a constant.

Halve the coefficient of `-6x`

`y=-1(x-3x)^2+8+k`

Discard the `x` from `-3x`

`y=-1(x-3)^2+8+k" "larr" nearly there. "k" is now fixed"`
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If you were to expand the bracket and compere to the original equation you would observe that there is an extra value. This is from squaring the `-3` from inside the bracket. `k` cancels that out.

Let `(-3)^2+k=0" "=>" "k=-9`

`color(brown)("However we have a -1 outside the bracket so this becomes.")`

Let `(-1)(-3)^2+k=0" "=>" "k=+9`

`color(brown)(y=-1(x-3)^2+8+k)color(blue)(->y=-1(x-3)^2+17`

Write as:`" "y=-(x-3)^2+17`