Question

Question #bac66

Answer 1

You have not asked for anything other than the equation form. Which is:

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

Explanation:

Given:`" "y=x^2-8x+18" ..........................Equation(1)"`

`color(brown)("Assumption: by 'graphing form' you mean completing the square")`

Let `k` be an error correction value.

Write as:

`y=(x^2-8x)+18 +k`

Take the power outside the brackets

`y=(x-8x)^2+18+k`

Halve the `8x`

`y=(x-4x)^2+18+k`

Remove the `x` from `4x`

`y=(x-4)^2+18+k" .....................Equation(2)"`
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`color(brown)("Explanation about dealing with the error")`
So we have now progressed from `ax^2+bx+c`

to `a(x+b/(2a))^2+c+k larr" equation(2) generalised"`

but in this case `a=1`

If we were to square out the brackets what we end up with would include `" "a(b/(2a))^2" "` which is the error so we have to set

`a(b/(2a))^2+k=0" "` to get rid of the error.
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In your question the error is `1xx(-4)^2`

Remember that `a=1` from `color(green)(color(red)(1x^2)-8x+18)`

So we set `(-4)^2+k=0 => k=-16`

Thus Equation(2) becomes

`y=(x-4)^2+18-16`

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