Question 1188441


{{{y = 2x^2 - 8x + 8}}}

At what {{{x}}}-value does the minimum occur? 

since you have a parabola that opens up, minimum is at vertex; so, complete square and write equation in vertex form

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

{{{y = 2(x^2 - 4x )+ 8}}}

{{{y = 2(x^2 - 4x +b^2)-2b^2+ 8}}}........{{{b=4/2=2}}}

{{{y = 2(x^2 - 4x +2^2)-2*2^2+ 8}}}

{{{y = 2(x- 2)^2-2*4+ 8}}}

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

{{{y = 2(x- 2)^2 }}}=> vertex is at ({{{2}}},{{{0}}})

so, minimum is {{{0}}} at {{{x=2}}}


domain: {{{R}}} (all real numbers)
range:{ {{{y}}} element {{{R}}} : {{{y>=0}}} } (all non-negative real numbers)


{{{ graph( 600, 600, -10, 10, -10, 10, 2x^2 - 8x + 8) }}}