# Question #bac66

Nov 10, 2016

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

$y = {\left(x - 4\right)}^{2} + 2$

#### Explanation:

Given:$\text{ "y=x^2-8x+18" ..........................Equation(1)}$

$\textcolor{b r o w n}{\text{Assumption: by 'graphing form' you mean completing the square}}$

Let $k$ be an error correction value.

Write as:

$y = \left({x}^{2} - 8 x\right) + 18 + k$

Take the power outside the brackets

$y = {\left(x - 8 x\right)}^{2} + 18 + k$

Halve the $8 x$

$y = {\left(x - 4 x\right)}^{2} + 18 + k$

Remove the $x$ from $4 x$

$y = {\left(x - 4\right)}^{2} + 18 + k \text{ .....................Equation(2)}$
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$\textcolor{b r o w n}{\text{Explanation about dealing with the error}}$
So we have now progressed from $a {x}^{2} + b x + c$

to $a {\left(x + \frac{b}{2 a}\right)}^{2} + c + k \leftarrow \text{ equation(2) generalised}$

but in this case $a = 1$

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

$a {\left(\frac{b}{2 a}\right)}^{2} + k = 0 \text{ }$ to get rid of the error.
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In your question the error is $1 \times {\left(- 4\right)}^{2}$

Remember that $a = 1$ from $\textcolor{g r e e n}{\textcolor{red}{1 {x}^{2}} - 8 x + 18}$

So we set ${\left(- 4\right)}^{2} + k = 0 \implies k = - 16$

Thus Equation(2) becomes

$y = {\left(x - 4\right)}^{2} + 18 - 16$

$y = {\left(x - 4\right)}^{2} + 2$ 