How do you write $f (x) = x^{2} - 2 x + 8$ $f(x) = x^2 - 2x + 8$ in vertex form?

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How do you write $f (x) = x^{2} - 2 x + 8$ $f(x) = x^2 - 2x + 8$ in vertex form?

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Answer 1 · 2016-08-15T10:47:13

In Vertex form $f (x) = {(x - 1)}^{2} + 7$ $f(x)=(x-1)^2+7$ .

Explanation:

$f (x) = x^{2} - 2 x + 8 = {(x - 1)}^{2} - 1 + 8 = {(x - 1)}^{2} + 7$ $f(x)=x^2-2x+8 =(x-1)^2-1+8=(x-1)^2+7$ .Vertex is at $(1, 7)$ $(1,7)$ graph{x^2-2x+8 [-40, 40, -20, 20]}[Ans]

Answer 2 · 2016-08-15T11:00:11

$f (x) = {(x - 1)}^{2} + 7$ $f(x) = (x-1)^2 +7" "$ Vertex is at $(1, 7)$ $(1, 7)$

Explanation:

Use the method of "completing the square."

$x^{2} - 2 x + 1 - 1 + 8 Add on half of (-2) squared$ $x^2 color(red)(- 2)x color(red)(+1) color(blue)(-1) +8" Add on half of (-2) squared"$

${(\frac{- 2}{2})}^{2} = {(- 1)}^{2} = 1$ $color(red)(((-2)/2)^2 = (-1)^2 = 1)$

But we may not change the value of the expression by just adding 1.
1 must be subtracted as well.
$+ 1 - 1 = 0$ $color(red)(+1)color(blue)( -1) = 0$

$x^{2} - 2 x + 1$ $x^2 -2x +1$ is a perfect square

$(x^{2} - 2 x + 1) - 1 + 8 = {(x - 1)}^{2} + 7$ $(x^2 -2x +1 ) -1 +8 = (x-1)^2 +7$

$f (x) = {(x - 1)}^{2} + 7 is the vertex form$ $f(x) = (x-1)^2 +7 " is the vertex form"$

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How do you write $f (x) = x^{2} - 2 x + 8$ $f(x) = x^2 - 2x + 8$ in vertex form?

2 Answers

Explanation:

Explanation:

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How do you write f(x)=x2−2x+8f(x) = x^2 - 2x + 8 in vertex form?

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How do you write $f (x) = x^{2} - 2 x + 8$ $f(x) = x^2 - 2x + 8$ in vertex form?