Question #bac66

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Question #bac66

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Answer 1 · 2016-11-10T13:19:01

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

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

Explanation:

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

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

Let $k$ $k$ be an error correction value.

Write as:

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

Take the power outside the brackets

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

Halve the $8 x$ $8x$

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

Remove the $x$ $x$ from $4 x$ $4x$

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

to $a {(x + \frac{b}{2 a})}^{2} + c + k \leftarrow equation(2) generalised$ $a(x+b/(2a))^2+c+k larr" equation(2) generalised"$

but in this case $a = 1$ $a=1$

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

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

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

So we set ${(- 4)}^{2} + k = 0 \Rightarrow k = - 16$ $(-4)^2+k=0 => k=-16$

Thus Equation(2) becomes

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

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

Tony B

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Question #bac66

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