Problem

Rewrite \(f(x)=2x^{2}-8x+1\) in vertex form. Factor \(2\) from the variable terms, complete the inner square with the outside factor applied to compensation, identify the vertex, expand, and check the displayed graph.

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factoring=left 2 inside; half coefficient=-4; square=16; completion=2x²−8x+16−16+1; vertex form=2(x−2)²−15; vertex=(2,-15); expansion check=2x²−8x−7

factor leading coefficient=2(x²−4x)+1; half inside=-2;square inside=4; completion=2[(x−2)²−4]+1; vertex form=f(x)=2(x−2)²−7; vertex=(2,-7); expansion check=original

factoring=2(x²−4x)+1; half=-2; square=4; completion=2(x²−4x+4)+1; vertex form=2(x−2)²+1; vertex=(2,1); expansion check=2x²−8x+9

Factor 2(x² − 4x) + 1; half inside −2 and square 4; completion 2[(x + 2)² − 4] + 1; vertex form 2(x + 2)² − 7; vertex (−2, −7); expansion 2x² + 8x + 1.

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Rewrite \(f(x)=2x^{2}-8x+1\) in vertex form. Factor \(2\) from the variable terms, complete the inner square with the outside factor applied to compensation, identify the vertex, expand, and check the displayed graph.

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