<PRE><B>Question 49923</B>
The vertex for y = ax^2 + bx +c will always be at {{{x=(-b)/(2a) }}}


In this case a=2 and b = -4, so vertex is at {{{x=4/(2*2)=1}}}


If x = 1, then find the y coordinate by substituting x= 1 into the equation for y, y = 2*1^2 - 4*1 = -2.  


This is a parabola that opens upward with vertex at (1,-2).  It should look like this:
{{{graph (300,300,-6,6,-6,6, 2x^2-4x) }}}


R^2 at SCC</PRE>