Question 1184797


{{{f(x) = 2x^2 - 8x + 1}}}.........complete square

{{{f(x) = (2x^2 - 8x) + 1}}}
{{{f(x) = 2(x^2 - 4x+b^2)-2b^2 + 1}}}..........{{{b=4/2=2}}}

{{{f(x) = 2(x^2 - 4x+2^2)-2*2^2 + 1}}}

{{{f(x) = 2(x - 2)^2-8 + 1}}}

{{{f(x) = 2(x - 2)^2-7}}}

vertex is at ({{{2}}},{{{-7}}})

the stationary point on the graph: 
The graph of a quadratic function (ie a parabola) only has a single stationary point.
For an ‘up’ parabola this is the minimum; for a ‘down’ parabola it is the maximum (no need to talk about ‘local’ here) The {{{y }}}value of the stationary point is thus the {{{minimum}}} or {{{maximum }}}value of the quadratic function

so,  the stationary point on the graph is {{{minimum}}} ({{{2}}},{{{-7}}})

or,

The specific nature of a stationary point at x can in some cases be determined by examining the second derivative {{{f}}}''{{{(x)}}}:
{{{f}}}'{{{(x)=2x-8}}}
{{{f}}}''{{{(x)=2}}}

{{{y=2*2^2 - 8*2 + 1}}}
{{{y= - 7}}}

the stationary point on the graph is  ({{{2}}},{{{-7}}})


{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(2,-7,.12),locate(2,-7,v(2,-7)),
graph( 600, 600, -10, 10, -10, 10,2x^2 - 8x + 1)) }}}