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

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Answer 1 · 2016-08-15T10:47:13

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

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

Answer 2 · 2016-08-15T11:00:11

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

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

$color(red)(((-2)/2)^2 = (-1)^2 = 1)$

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

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

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

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

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