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"#