What is the vertex form of $y = x^{2} - 2 x + 8$ $y=x^2-2x+8$ ?

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Answer 1 · 2018-03-24T11:28:42

Vertex form of equation is $y = {(x - 1)}^{2} + 7$ $y =(x-1)^2 +7$

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

$y = {(x - 1)}^{2} + 7$ $y=(x-1)^2 +7$ Comparing with vertex form of

equation $f (x) = a {(x - h)}^{2} + k; (h, k)$ $f(x) = a(x-h)^2+k ; (h,k)$ being vertex we find

here $h=1 , k=7, a=1 :.$ Vertex is at $(1,7)$

and vertex form of equation is $y =(x-1)^2 +7$

graph{x^2-2x+8 [-35.54, 35.58, -17.78, 17.78]} [Ans]

What is the vertex form of y=x^2-2x+8y=x2−2x+8y=x^2-2x+8?